Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Inferring observation time from a Brownian motion This might be a bit lengthy question. So let me proceed in steps.
General description: I have some observations, based on which I want to infer their occurring time.
Specific setting: Let $W(t)$ be a Brownian motion.
I have an observation $X$ defined as
$$X = BW(\delta) + (1-B)W(2\delta),$$
where $B$ is a Bernoulli draw with success rate $\frac{1}{2}$ and is independent of $W(t)$; and $\delta>0$ is some constant observation latency.
In words, the above is meant to capture the setting that I do not know when the observation $X$ occurred:
It might have occurred at time $t=\delta$ or at time $t=2\delta$, equally likely.
So given the observation $X$, I can infer something about the time of its occurrence: By Bayesian rule,
$$\mathbb{P}(B=1|X) = \frac{\mathbb{P}(B=1)\mathbb{P}(X|B=1)}{\mathbb{P}(B=1)\mathbb{P}(X|B=1)+\mathbb{P}(B=0)\mathbb{P}(X|B=0)}
= \frac{\frac{1}{2}\mathbb{P}(W(\delta)=X)}{\frac{1}{2}\mathbb{P}(W(\delta)=X)+\frac{1}{2}\mathbb{P}(W(2\delta)=X)}.
$$
Given that $W(t)$ is a Brownian motion, $W(\delta)$ and $W(2\delta)$ are normally distributed with respective densities
$$\mathbb{P}(W(\delta)=X) = \frac{1}{\sqrt{2\pi\delta}}e^{-X^2/(2\delta)}$$
and
$$\mathbb{P}(W(2\delta)=X) = \frac{1}{\sqrt{4\pi\delta}}e^{-X^2/(4\delta)}.$$
So the above translates to a function of the density ratio
$$\mathbb{P}(B=1|X) = \frac{1}{1+\frac{1}{\sqrt{2}}e^{X^2/(4\delta)}},
$$
which is decreasing in the magnitude of $|X|$.
That is, the more extreme $X$ realizes to be, the less likely it comes from the early $W(\delta)$.
This makes sense as $W(\delta)$ is less volatile than $W(2\delta)$.
Next I want to consider what happens when $\delta\downarrow0$, i.e., when the gap between the timing of the observation asymptotically vanishes.
Intuitively, when $\delta\downarrow0$, $W(\delta)=W(2\delta)=W(0)$ and $X=W(0)$, leaving no room for inference for $B$.
Analytically, however, from the above expression, in the limit of $\delta\downarrow0$, $\mathbb{P}(B=1|X)=0$, suggesting that I should always infer that $B=0$.
So what is the correct result? Can I infer anything about $B$ in the limit of $\delta\downarrow0$?
| The probability $$\mathbb{P}(B = 1 \mid X) = \frac{1}{1 + \frac{1}{\sqrt{2}}e^{X^2/(4\delta)}}$$ only depends on $X$ and $\delta$ through the term $X^2/(4\delta)$. Now let's think about the cases when $B = 1$ or $B = 0$.
When $B = 1$ then $X \sim N(0, \delta)$, so $X/\sqrt{\delta} \sim N(0, 1)$. Therefore $X^2/\delta \sim \chi^2_1$, a chi-squared distribution. And $X^2/(4\delta) \sim \chi^2_1/4$, the distribution of a $\chi^2_1$ random variable divided by $4$.
When $B = 0$ then $X \sim N(0, 2\delta)$, so $X/\sqrt{2\delta} \sim N(0, 1)$. Therefore $X^2/(2\delta) \sim \chi^2_1$, and $X^2/(4\delta) \sim \chi^2_1/2$, the distribution of a $\chi^2_1$ random variable divided by $2$.
In either case, the distribution of $X^2/(4\delta)$ does not depend on $\delta$. So as $\delta \rightarrow 0$, $\mathbb{P}(B = 1\mid X)$ does not tend to any particular limit.
| {
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"timestamp": "2023-03-29T00:00:00",
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proof about injection I have to proof that the function $f : X \rightarrow Y$ is an injection if and only if $\forall T \subseteq X$, $f(X\setminus T) \subseteq Y \setminus f(T)$.
I'm having some difficulties. First (1) I proof that if $f$ is an injection then $\forall T \subseteq X$, $f(X\setminus T) \subseteq Y \setminus f(T)$, successively I'll prove the inverse implication (2).
(1):
I want to show that a generic element of $f(X\setminus T)$ belongs to $Y \setminus f(T)$ too, but I don't know how to continue, it is not apparent to me how to use the injection of $f$.
(2)
EDIT as suggested by Mark, i'll try the direction 2.
We know that $\forall x \in X\setminus T$ and $\forall t \in T$, we have $x\neq t $, because x belongs to X but it does not belong to T. Now , if I choose $y \in f(X \setminus T)$, there is $x \in X \setminus T$ : $ y = f(x)$, because y is an element of the image of $X\setminus T$ through $f$. But we know also that $y \in Y\setminus f(T)$, so $y \notin f(T)$, this implies that $\forall t \in T$, $y\neq f(t)$. This reduces to $f(t) \neq f(x) $. Is this proof valid ?
| So we want to show $f$ is injective iff
$$\forall T \subseteq X: f[X\setminus T]\subseteq Y \setminus f[T]\tag{*}$$
So let $f$ be injective, $T \subseteq X$ and let $y \in f[X\setminus T]$, i.e.
$y=f(x)$ with $x \notin T$. By definition of the function, $y \in Y$, but I claim that also $y \notin f[T]$, because otherwise $y=f(t)$ for some $t \in T$, and as $x \notin T$, we have that $x \neq t$ but also $y=f(x)=f(t)$ contradicting that $f$ is injective. So $y \in Y\setminus f[T]$, and $(\ast)$ has been shown.
Suppose that $(\ast)$ holds. Suppose $f$ were not injective, so that we have that $x_1 \neq x_2$ in $X$ with $y:= f(x_1) = f(x_2)$. Then $f[X\setminus \{x_1\}] = Y$ (as the only point we could miss is $y$ but this is also assumed by $x_2 \in X\setminus \{x_1\}$) but $Y\setminus \{y\}$ is a proper subset of $Y$, so that $(\ast)$ fails for $T=\{x_1\}$ (or $T=\{x_2\}$ too).
This contradiction shows that $f$ is injective.
| {
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A bag contains 2 red, 3 green and 5 blue balls A ball is drawn, observed and put again in the bag. Find the probability of getting all colours different.
My teacher's solution:
2/10 × 3/10 × 5/10 × 3! × 2
I don't get why that last 2 was necessary
Edit: balls are drawn 3 times
| The last $2$ should not be there. The first three terms give the chance of getting specifically red,green,blue. The $3!$ is the number of orders of colors and you are done.
| {
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Proof: For any integer $n$ with $n \ge 1$, the number of permutations of a set with $n$ elements is $n!$. I am trying to use mathematical induction to prove the following theorem:
For any integer $n$ with $n \ge 1$, the number of permutations of a set with $n$ elements is $n!$.
Proof
Let $P(n)$ be the above statement.
Take the set of elements $\{ x_1, x_2, \dots, x_n \mid n \ge 1 \}$.
$P(1)$ holds because the number of permutations of $1$ element is size $1$ and $1! = 1$.
Now assume that $P(n)$ is true for some $n = m \ge 1$.
$P(m + 1)$ means that we have the set $\{ x_1, x_2, \dots, x_{m + 1} \}.$
I'm not sure how to proceed from here. I was thinking of using the multiplication rule somehow, but I've been unable to make progress on this.
I've also been unable to find any proofs for this theorem online.
I would greatly appreciate it if people could please help me prove this.
EDIT (Completed Proof):
Let $P(n)$ be the above statement.
Take the set $X = \{ x_1, x_2, \dots, x_n \mid n \ge 1 \}$.
$P(1)$ holds because the number of permutations of $1$ element is size $1$ and $1! = 1$.
Now assume that $P(n)$ is true for some $n = m \ge 1$.
And assume we have the sets $X = \{ x_1, x_2, \dots, x_m \}$ and $X' = \{ x_{m + 1} \}$.
Let task $T$ represent tasks $T_1, T_2, \dots, T_m$, where task $T_k, k = 1, 2, \dots, m$, represents the task where the $k$th element of the set $X$ is fixed and every permutation of the resultant set configuration is calculated.
Every time we fix one element and find all permutations of the resultant set, that leaves one set configuration that the next set cannot have since it would be identical to one of the permutations of the previous configuration. This is what we assumed to be true for a set with $m$ elements.
Let task $T_{m + 1}$ be the task where we take the set $X^* = X \cup X'$, fix the element $x_{m + 1}$, and calculate all permutations of the resultant set configuration. Since there are $m + 1$ elements in the set $X^*$, there are $m + 1$ ways to fix $x_{m + 1}$ ($m + 1$ set configurations) and calculate all permutations. Therefore, according to the multiplication rule, there are $(m!)(m + 1) = (m + 1)!$ ways to perform tasks $T$ and $T_{m + 1}$. $$\tag*{$\blacksquare$}$$
I would greatly appreciate it if people could please review the complete proof and provide feedback as to its correctness.
| When we are given $m+1$ items, we separate a special item from the other $m$ items. We sort the $m$ items and fix their order, that give us $m!$ options.
Now we have $m+1$ positions to choose to place our special items and none of them repeats.
Hence we have $m! \cdot (m+1)= (m+1)!$
| {
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"timestamp": "2023-03-29T00:00:00",
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Roots Across the Complex Numbers Why is it the case that an even root (square root, quartic, etc) can be positive or negative across the complex numbers, but is limited to postive in the reals? Is there a good mathematical reason for this, or is it simply notation?
| Essentially, the reason is that complex multiplication, and by extension exponentiation, are far more complicated than in the reals. There is a rotational aspect, and for an $n$th root, there are $n$ answers. It would be silly to declare that only one of these is the "actual" answer, whereas in the reals it makes much more sense. Glad to see that you have joined the site. Please continue asking questions.
| {
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Let $G$ be transitive on $S$. Show that the action is primitive if and only if every $\operatorname{Stab}_G(a), a\in S$, is a maximal subgroup of $G$. I am self-studying "Classical Groups and Geometric Algebra" by Larry C. Grove. This is the 2nd question of the exercises of the 0th Chapter.
Let $G$ be transitive on $S$. Show that the action is primitive if and only if every $\operatorname{Stab}_G(a), a\in S$, is a maximal subgroup of $G$.
Unfortunately, I can't even start.
*
*$G$ acts transitively on $S$ $\iff$ $\operatorname{Orb}_G(a)=S, \forall a\in S$.
So, we can reach any element of $S$ applying $G$ to any element of $S$.
*If the action is primitive there is no block like $B\subseteq S$,
with $|B|\ge 2, B\neq S$, such that for each $x \in G$ either $xB=B$
or $xB\cap B = \emptyset$.
*$\operatorname{Stab}_G(a)$ is a maximal subgroup if there is no subgroup containing
$\operatorname{Stab}_G(a)$ other than $\operatorname{Stab}_G(a)$ itself and $G$.
| ($\implies$)
Let $H$ is a proper subgroup of $G$ containing $Stab_G(a)$ and $B=\{ha|h\in H\}$ is a subset of $S$.
Assume that $xB\cap B \neq \emptyset$ for some $x \in G$. There exist $h,h'\in H$ such that $xha=h'a$. Then, $h'^{-1}xha=a\implies h'^{-1}xh\in Stab_G(a)\implies h'^{-1}xh\in H\implies h'(h'^{-1}xh)h^{-1}=x\in H\implies xB=B\implies B\text{ is a block.}$
$B$ is a block means $B=\{a\}$ or $B=S$ because $G$ acts primitively on $S$.
*
*$B=\{ha|h\in H\}=\{a\} \implies ha=a, \ \forall h\in H \implies H=Stab_G(a)$.
*$B=\{ha|h\in H\}=S \implies H\text{ acts transitively on }B.$ We also have $G$ acts transitively on $B$. So there exist $g\in G$ and $\bar h\in H$ such that $ga=\bar ha$. Then, $\bar{h}^{-1}ga=a\implies \bar{h}^{-1}g\in H\implies G\subseteq H\implies H=G$.
($\impliedby$)
Let $G$ acts on $S$ imprimitively such that there exists some block $B\subseteq S$ with $a\in B$. Then, $Stab_G(B)=\{g:gb=b, \ g\in G, \ and \ \forall b\in B\}\supset\{g:ga=a, \ g\in G \}=Stab_G(a)$. Let $\alpha, \ \beta\in Stab_G(B)$, and $b\in B$. Then $(\alpha\beta)\cdot b=\alpha\cdot(\beta\cdot b)=\alpha\cdot b=b\implies \alpha\beta\in Stab_G(B).$ Moreover, $\alpha^{-1}\cdot b=\alpha^{-1}\cdot(\alpha\cdot b)=(\alpha^{-1}\alpha)\cdot b=b\implies \alpha^{-1}\in Stab_G(B).$ Thus, $Stab_G(B)$ is a proper subgroup containing $Stab_G(a)$.
| {
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Number of compositions of $n$ such that each term is less than equal to $k.$ Let $n$ be an integer $\geq 1.$ Then a partition of $n$ is a sequence of positive integers (greater than equal to $1$) such that their sum equals $n.$ So for instance if $n=4$ then
$$[[4], [1, 3], [1, 1, 2], [1, 1, 1, 1], [1, 2, 1], [2, 2], [2, 1, 1], [3, 1]]$$
is a collection of all partitions of $n$ with order. Clearly for any $n$ the number of such ordered partitions is $2^{n-1}.$ However I want to count the number of paritions of $n$ where each integer in the parition is less than equal to some integer $k.$ So for instance if $n=4$ and $k=2$ then
$$[[1, 1, 2], [1, 1, 1, 1], [1, 2, 1], [2, 2], [2, 1, 1]]$$
is a collection of all the $2$-partitions of $4.$ Is there a general formula for finding this or maybe an asymptotic expression?
| First a note about terminology:
- a Partition of a positive integer $n$ is a non-decreasing sequence of positive integers summing to $n$;
- a Composition of a positive integer $n$ is an unordered sequence of positive integers summing to $n$.
That premised, you are speaking of the number of Compositions of $n$, whose terms (parts) are not-greater than $k$.
Consider the case in which you seek for the number of compositions of the positive number $s+m$
into $m$ parts not exceeding $r+1$
$$
{\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
{\rm 1} \le {\rm integer}\;y_{\,j} \le r + 1 \hfill \cr
y_{\,1} + y_{\,2} + \; \cdots \; + y_{\,m} = s + m \hfill \cr} \right.
$$
that's the same as
$$N_{\,b} (s,r,m) = \text{No}\text{. of solutions to}\;\left\{ \begin{gathered}
0 \leqslant \text{integer }x_{\,j} \leqslant r \hfill \\
x_{\,1} + x_{\,2} + \cdots + x_{\,m} = s \hfill \\
\end{gathered} \right.$$
$N_b$ is given by the following sum
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,j\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)}
{\left( { - 1} \right)^k \binom{m}{j}
\binom
{ s + m - 1 - j\left( {r + 1} \right) }
{ s - j\left( {r + 1} \right)}\ }
$$
as widely explained in this related post.
Going back to your case, the
number of compositions of $n$ into $m$ parts not greater than $k$
will then be
$$
\eqalign{
& N_c (n,k,m)\quad \left| {\;1 \le n,m,k} \right.\quad = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,\,j\,\,\left( {\, \le \,m} \right)} {\left( { - 1} \right)^{\,j} \binom{m}{j}
\binom{n - 1 - j\,k}{ n - m - j\,k}} \cr}
$$
while the
overall number of compositions of $n$ into parts not greater than $k$
will be the sum
of the above for $0 \le m$ : if the sum extends over $n$ it will maintain the value $2^{n-1}$.
$$ \bbox[lightyellow] {
\eqalign{
& N_{c\,t} (n,k)\quad \left| {\;0 \le n,k \in \mathbb Z} \right.\quad = \cr
& = \sum\limits_{0\, \le \,\,m} {\;\sum\limits_{\left( {0\, \le } \right)\,\,j\,\,\left( {\, \le \,m} \right)} {\left( { - 1} \right)^{\,j}
\binom{m}{j} \binom{ n - 1 - j\,k}{n - m - j\,k}
} } \cr}
}$$
In the example with $n=4$, the above gives for $k=1,2,3,4$
$$
1,5,7,8
$$
which in fact correspond to
$$
\eqalign{
& \left[ {1,1,1,1} \right] \cr
& \left[ {1,1,1,1} \right]\left[ {1,1,2} \right]\left[ {1,2,1} \right]\left[ {2,1,1} \right]\left[ {2,2} \right] \cr
& \left[ {1,1,1,1} \right]\left[ {1,1,2} \right]\left[ {1,2,1} \right]\left[ {2,1,1} \right]\left[ {2,2} \right]\left[ {1,3} \right]\left[ {3,1} \right] \cr
& \left[ {1,1,1,1} \right]\left[ {1,1,2} \right]\left[ {1,2,1} \right]\left[ {2,1,1} \right]\left[ {2,2} \right]\left[ {1,3} \right]\left[ {3,1} \right]\left[ 4 \right] \cr}
$$
Finally note that:
- $N_{c\,t}(n,k)$ correctly checks with OEIS seq. A126198, which provides further properties of these numbers;
- the o.g.f. of $N_{c\,t}$ in $n$ is in fact the $F(x)$ given by @jmerry's answer (re. to the o.g.f. for $N_b$ given in related post);
$$
\sum\limits_{0\, \le \,\,n} {N_{c\,t} (n,k)\,x^{\,n} } = {{1 - x} \over {1 - 2x + x^{\,k + 1} }}
$$
- $N_{c\,t}(n,k)$ satisfies the recursion given by @NobleMushtak
$$
N_{c\,t} (n,k) = \sum\limits_{j = 1}^k {N_{c\,t} (n - j,k)} + \left[ {n = 0} \right]
$$
where $[P]$ denotes the Iverson bracket.
| {
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find the value of normal distribution with a 3 decimal places Z score using table In a past exam of my stats class, the question requires finding the value of a normal CDF corresponding to a Z score of 3 DP. However, we are only given a table that is accurate to 2 DP. We are not allowed to use calculator. What should I do to find the value when the Z score is 1.293, not just 1.29.
| Personally, I would just round to $1.29$. However, if you really want to guess $\text{normalcdf}(1.293)$, I would just do a weighted sum.
First, write $1.293$ in terms of $1.29$ and $1.30$, since those are the two-decimal z-scores closest to it:
$$1.293=0.7\cdot 1.29+0.3\cdot 1.30$$
Then, pretend $\text{normalcdf}$ is linear:
$$\text{normalcdf}(1.293)=0.7\text{normalcdf}(1.29)+0.3\text{normalcdf}(1.30) \\ =0.7\cdot 0.9015+0.3\cdot 0.9032=0.90201$$
| {
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Why do I keep getting this incorrect solution when trying to find all the real solutions for $\sqrt{2x-3}\ +x=3$. The problem is to find all real solutions (if any exists) for $\sqrt{2x-3}\ +x=3$.
Now, my textbook says the answer is {2}, however, I keep getting {2, 6}. I've tried multiple approaches, but here is one of them:
I got rid of the root by squaring both sides,
$$\sqrt{2x-3}^2=(3-x)^2$$
$$0=12-8x+x^2$$
Using the AC method, I got
$$(-x^2+6x)(2x-12)=0$$
$$-x(x-6)2(x-6)=0$$
$$(-x+2)(x-6)=0$$
hence, $$x=2, \ x=6$$
Of course, I can always just check my solutions and I'd immediately recognize 6 does not work. But that's a bit too tedious for my taste. Can anyone explain where I went wrong with my approach?
| Because squaring both sides of an equation always introduces the “risk” of an extraneous solution.
As a very simple example, notice the following two equations:
$$x = \sqrt 4 \iff x = +2$$
$$x^2 = 4 \iff \vert x\vert = 2 \iff x = \pm 2$$
The first equation has only one solution: $+\sqrt 4$. The second, however, has two solutions: $\pm\sqrt 4$. And you get the second equation by squaring the first one.
The exact same idea applies to your example. You have
$$\sqrt{2x-3} = 3-x$$
which refers only to the non-negative square root of $2x-3$. So, if a solution makes the LHS negative, it is extraneous. But, when you square both sides, you’re actually solving
$$0 = 12-8x+x^2 \iff \color{blue}{\pm}\sqrt{2x-3} = 3-x$$
which has a $\pm$ sign and is therefore not the same equation. Now, to be precise, you’d have to add the condition that the LHS must be non-negative:
$$2x-3 = 9-6x+x^2; \quad \color{blue}{x \leq 3}$$
$$0 = 12-8x+x^2; \quad \color{blue}{x \leq 3}$$
Now, your equation is equivalent to the first with the given constraint. If you get any solution greater than $3$, (in this case, $6$), you’d know it satisfies the new equation but not the original one.
| {
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Probability of getting 6 heads in a row from 200 flips and intuition about this high value A few days ago I had an argument with a friend about this question :
What is the probability of getting 6 heads in a row from 200 flips ?
I argued it is high probability (significantly bigger than a half) while he argued it is low probability.
When I tried to give exact formula I failed, so we checked the web were the answer was about 84%, yet he is still not convinced so from this I have two questions:
*
*What is the exact formula for $k$ Heads in a row (consecutive) out of $n$ coin flips?
*(Not a mathematical) How to convince my friend that 6 in a row have high probability? Meaning what is the intuition behind the question ?
| Here is how to calculate the exact answer. Consider a Markov chain $X_0,X_1,\ldots,X_{200}$, taking integer values in the range $0\le X_n\le 6$, with
transition matrix (with row and column indices in the range $0\le i,j\le6$) $$M=\pmatrix{\frac12&\frac12&0&0&0&0&0&\\
\frac12&0&\frac12&0&0&0&0&\\
\frac12&0&0&\frac12&0&0&0&\\
\frac12&0&0&0&\frac12&0&0&\\
\frac12&0&0&0&0&\frac12&0\\
\frac12&0&0&0&0&0&\frac12\\
0&0&0&0&0&0&1}$$
Here the idea is that $X_n$ represents the number of consecutive heads ending at flip $n$ (with the conventional courtesy value $X_0=0$) so that the flip sequence HTHH would cause $X_0=0$, $X_1=1$, $X_2=0$, $X_3=1$, $X_4=2$, and so on. Except the value $X_n=6$ means something different: either $X_{n-1}=6$ or the $n$-th flip was H and $X_{n-1}=5$. The chain is started with the value $X_0=0$; what is sought is the probability that $X_{200}=6$. This is the $(0,6)$-th entry in the matrix $M^{200}$. When I do these calculations I get a value very close to $.8$.
Here is a way to visualize this chain. There is a small spider that aspires to climb to the top of a $6$ segment pipe. It starts at the bottom, and does the following $200$ times:
*
*if it is at the top, it stays at the top,
*otherwise it flips a coin and drops all the way to the bottom if the coin shows T, and climbs up more more segment if the coin shows H.
| {
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How to solve this Diophantine equation? Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.
Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...
| "OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
| {
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I cannot calculate $\tan^{-1}(1+i)$ I use the formula below for inverse tangent function:
$$\tan^{-1}(z)= \frac {i} 2 \log \biggr( \frac {i+z}{i-z} \biggr)$$
I have written
$$\tan^{-1}(1+i)= \frac {i} 2 \log \biggr( \frac {i+1+i}{i-1-i} \biggr)= \frac {i} 2 \log (-2i-1)$$
The answer is $n\pi i , n\in \mathbb Z$ but I don’t know or I cannot see what is the argument of $-2i-1$
I know it is basic question but I couldn’t see how can I do
Thanks for helps
| The number $-1-2i$ is in the third quadrant, so its argument, taken in $[0,2\pi)$, is
$$
\theta=\pi+\arctan\frac{-2}{-1}=\pi+\arctan2
$$
(it would be wrong to say that the argument is $\arctan2$). Thus
$$
-1-2i=\sqrt{5}e^{i\theta}
$$
and its logarithms are
$$
\frac{1}{2}\log5+i\theta+2k\pi i
$$
and therefore
$$
\frac{i}{2}\left(\frac{1}{2}\log5+i\theta+2k\pi i\right)=-\theta-k\pi + \frac{i}{4}\log5=
-\frac{\pi}{2}-\frac{1}{2}\arctan2-k\pi + \frac{i}{4}\log5
$$
that can also be written as
$$
\frac{\pi}{2}-\frac{1}{2}\arctan2+k\pi+\frac{i}{4}\log5
$$
| {
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In how many ways can I write $0$ as a sum of $n\; 0s, 1s \;\text{and}\; -1s?$ In how many ways can I write $0$ as a sum of $n\; 0s, 1s \;\text{or}\; -1s?$ (Taking the order into account).
I suspect there is no closed formula to express the result, but I'd like someone to confirm it, or deny it.
Edit:
e.g., if $n=3$
$$\begin{aligned}0&=\;\;\;0+0+0,\\ 0&=\;\;\;1-1+0,\\ 0&=\;\;\;1+0-1,\\ 0&=\;\;\;0+1-1,\\ 0&=-1+1+0,\\ 0&=-1+0+1,\\ 0&=\;\;\;0-1+1.\end{aligned}$$
So for $n=3\;$ there are $7$ ways.
| The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:
$$2p+q=n| p,q \in \Bbb Z^+$$
This can be done in $\frac{n+1}{2}$ ways for odd $n$ and $\frac{n+2}{2}$ ways for even $n$
You'll just need to account for positioning after this.
| {
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Meaning of "$\Leftrightarrow$" I just read (at the beginning of this Wikipedia site) that "$A \Leftrightarrow B$" means "A can be replaced in a logical proof with B". Is this interpretation of the symbol true?
(I have so far understood the symbol for equivalence ($\Leftrightarrow$) to mean that, given $A \Leftrightarrow B$, then if A is true then B is true and vice versa and, if A is false then B is false and vice versa.)
| $$\iff \text { means *if and only if* }$$
Essentially we have that:
$$A\iff B \to (A \implies B) \text { AND } (B\implies A)$$
"A if and only if B is the same as A if B and B if A"
| {
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For which $a$ does the equation $a^x=x+2$ have two solutions? I need to find values of $a$ for the following equation to have two real solutions.
$$a^x=x+2$$
*
*$(1,\infty)$
*$(0,1)$
*$1/e,e$
*$(1/(e^e), e^e)$
*$(e^{1/e}, \infty)$
This is how I solved this exercise, but I don't understand some things.
I would like to know if there's another way to solve this kind of exercise. I would be happy if I would get some ideas.
Also, from my solution, I don't understand why from that table results just one solution and from the graphic results two solutions. Usually, to see the number of solutions I use this kind of table.
For $a>1$, $f$ decreases from infinity to -1, then increase from -1 to infinity. I'm really confused. Need some suggestions here.
Thank you!
| Your attempt is wrong, sorry: you cannot just use particular cases. And the case $a=-e$ is impossible, because $a^x$ is only defined for $a>0$. The answer should be in terms of $a$, and using a single value is not enough.
Consider the function $f(x)=a^x-x-2$. Then
$$
f'(x)=a^x\log a-1
$$
(with $\log$ being the natural logarithm).
This doesn't vanish for $0<a\le 1$, so the function can have two zeros only for $a>1$.
In this case the point of minimum is at
$$
x=-\frac{\log\log a}{\log a}
$$
Set $b=\log a$, for simplicity. Then $a=e^b$ and $a^x=e^{bx}$; we want to evaluate
$$
f\left(-\frac{\log b}{b}\right)=e^{-\log b}+\frac{\log b}{b}-2=\frac{-1+\log b-2b}{b}
$$
Consider $g(t)=-1+\log t-2t$, for $t>0$; then $g'(t)=\frac{1}{t}-2$, which vanishes for $t=1/2$; since
$$
g(1/2)=-1-\log2-1<0
$$
you have the desired answer, because this implies $g(b)<0$.
| {
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Question related to Darboux's theorem Darboux's theorem says that $f'$ has intermediate value property. More precisely,
( Darboux's theorem) If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then $\exists$ c in (a,b) such that $f'(r)=c$.
Thus Darboux's theorem implies that $f'$ cannot have any simple discontinuities on $[a,b].$
I have a question as follows:
Is there a function $f$ satisfying $\displaystyle \lim_{x\to d} f'(x)=\infty$ for some $d \in (a,b)$ under the assumptions in Darboux's theorem?
If so, I can conclude that for each $x \in (a,b),$ either $(i)$ $f'$ is continuous at $x$ or $(ii)$ $f'$ oscillates near $x.$
It seems that there is no differentiable function $f$ on $[a,b]$ satisfying function $\displaystyle \lim_{x\to d} f'(x)=\infty$ and $\exists f'(d)$ for some $d \in (a,b)$.
Please let me know if you have any idea or comment for my question. Thanks in advance!
| No, there can't be.
If $f'(x)\to +\infty$ for $x\to d$, then there will be $\delta$ such that $f'(x) > f'(d)+2$ on $(d, d+\delta)$. And this contradicts Darboux's theorem on $[d,d+\delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.
Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.
| {
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Commutative local ring with $10$ ideals Let $R$ be a commutative ring with unity with exactly $10$ ideals (including $\{0\}$ and $R$ ) . Then is it true that $R$ is a Principal Ideal Ring ?
My Work: I know that any commutative ring with $5$ or less ideals is a PIR. Indeed, suppose $R$ has $5$ or less ideals. Then $R$ is Noetherian so every ideal is finitely generated. If $R$ is not PIR, then there exists a $2$-generated ideal which is not principal, call it $J=(a,b)$. Then $(0);(a); (b); ( a+b); (a,b) $ and $R$ are all distinct ideals of $R$ giving at least $6$ ideals, contradiction ! Hence any ring with $5$ or less ideals is PIR. Now using this it follows that if $R$ is not local and has $10$ ideals, then by structure thm of Artinian rings, $R\cong R_1 \times R_2$ , where $R_1, R_2$ are non-zero Artinian rings having say $n_1$ and $n_2$ many ideals respectively. Then $n_1n_2=10$. Since $n_1,n_2 \ge 2$, we get w.l.o.g. $n_1=2, n_2=5$, thus both $R_1,R_2$ are PIR, hence $R\cong R_1 \times R_2$ is a PIR.
Thus, we only need to check the case for local rings.
| I count exactly ten ideals in $R=\mathbb{F}_5[x,y]/(x^2,y^2)$. Namely
$$0,(xy),(x),(x+y),(x+2y),(x+3y),(x+4y),(y),(x,y),R.$$
And $(x,y)$ is not principal, so $R$ is not a PIR.
| {
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Find the $x$ at which the local maxima of a function ocours.
The function $f(x) = \int\limits_{-1}^{x}t(e^t-1)(t-1)(t-2)^3(t-3)^5 dt$ has a local maxima at $x=?$
First, I differentiated $f(x)$ and found its roots.
That came out to be $x = 0,1,2,3$. Now, one of those numbers, when plugged into $f(x)$, must give the largest value compared to the others.
In our case, since we are integrating, we would need to find the $x$ that minimizes the negative area of the function that we are integrating.
By further work and analyzing the function, I figured out that the answer must be $0$ or $2.$
But, I don't know what to do next?
Any help would be appreciated.
| HINT:
The sign of $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5$$ is $-,+,-,+$ respectively on the intervals $(-1,1),(1,2),(2,3),(3,\infty).$ From this you can finish without further calculation.
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How is it possible that $\textrm{HK}=\textrm{G}$? For the following problem:
If $\textrm{H}$ and $\textrm{K}$ are distinct subgroups of $\textrm{G}$ of index $2$, then $\textrm{H}\cap\textrm{K}$ is a normal subgroup of $\textrm{G}$ of index $4$.
It is obvious that both $\textrm{H}$ and $\textrm{K}$ are normal subgroups since both have index $2$. Hence their intersection $\textrm{H}\cap\textrm{K}$ is also normal.
The main issue comes with proving that it has an index $4$, using 2nd Isomorphism Theorem. Clearly $\displaystyle\frac{\textrm{H}}{\textrm{H}\cap\textrm{K}}$ is isomorphic to $\displaystyle\frac{\textrm{HK}}{\textrm{K}}$ and using this we can show that $\textrm{H}\cap\textrm{K}$ that index $4$. But on this forum, I have seen another proof of this problem where a person wrote that $\textrm{H}\textrm{K} = \textrm{G}$.
How the heck are they equal? I'm having extreme difficulty in proving this.
Has this got something to do with both $\textrm{H}$ and $\textrm{K}$ being distinct ?
| It has to do with $H$ and $K$ being distinct and having index $2$: these hypotheses imply that none of them is contained in the other. So take an element $h\in H\smallsetminus K$. Then $\;G=hK\cup K$ since $K$ has index $2$. Also, since both $hK,K\subset HK$, we have $G=HK$.
| {
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How do I get from $x - x^2 = \frac{1}{4}$ to $x =\frac{1}{2}$? I'm working on a text book problem where I need to sketch the graph of $y = 4x^2 - 4x+1$ by finding where the curve meets the $x$ axis.
To start out I set $y = 0$ then tried to isolate $x$ then,
$4x - 4x^2 = 1$
$x - x^2 = \frac{1}{4}$
From here I want to continue algebraically to reach $x = \frac{1}{2}$ which I know is the solution from plotting the curve using an app and I can see that that makes sense.
However I am missing some concepts which allow me to turn $x - x^2 = \frac{1}{4}$ into $x = \frac{1}{2}$ and wanted some help to get unblocked.
| $$4x-4x^2=1\implies 4x^2-4x+1=0\implies (2x-1)^2=0\implies 2x-1=0$$
| {
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How did Einstein integrate $\frac{\partial \tau}{\partial x'}+\frac{v}{c^2-v^2}\frac{\partial \tau}{\partial t}=0$?
In his paper "On the Electrodynamics of Moving Bodies", Einstein writes the equation
$$\dfrac{\partial \tau}{\partial x'}+\dfrac{v}{c^2-v^2}\dfrac{\partial \tau}{\partial t}=0$$
where
*
*$\tau=\tau(x',y,z,t)$ is a linear function (i.e. $\tau=Ax'+By+Cz+Dt$)
*$x'=x-vt$
*$\dfrac{\partial \tau}{\partial y}=0$ (i.e. $B=0$)
*$\dfrac{\partial \tau}{\partial z}=0$ (i.e. $C=0$)
*$c$ is a constant
*$x,x',y,z,t,v$ are variables
and he derives that
$$\tau=a\left(t-\dfrac{v}{c^2-v^2}x'\right)$$
where $a=a(v)$
Could someone please walk me through step-by-step how he derived this? I am not very familiar with integrals invovling partial derivatives, so I would be appreciative if any answers are quite explicit.
Additionally, if I modified the question to say that $\tau$ was an affine function (i.e. $\tau=Ax'+By+Cz+Dt+E$), would it make any difference to the result (I suspect it wouldn't)?
| From the definitions given:
$$\partial_{x'}\tau = A, \quad \partial_t \tau = D$$
Also:
$$\partial_{y}\tau = B = 0, \quad \partial_z \tau = C = 0$$
From the differential equation we get:
$$A + \frac{v}{c^2 - v^2}D = 0 \\
\implies A = - \frac{v}{c^2 - v^2}D$$
Now using the definition given for $\tau$:
$$\tau = - \frac{v}{c^2 - v^2}Dx' + Dt = D\bigg(t - \frac{v}{c^2 - v^2}x'\bigg)$$
So if you define $D=a$ you get the final expression for $\tau$. As you have noticed, requiring $\tau$ to be affine doesn't change the results at all - your $\tau$ would be:
$$ \tau = a\bigg(t - \frac{v}{c^2 - v^2}x'\bigg) + E$$
The reason $a=a(v)$ is because $D$, who is actually $a$ in disguise, does not depend on $x', y, z, t$ but is assumed to depend on $v$. Otherwise, the relation wouldn't be linear. Without further information, $a$ is a function of potentially anything except those four variables.
| {
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Affine transformations satisfying conditions I'm asked to find ALL affine transformations from $\mathbb{R^3}$ to itself which satisfy that the point $(-1,2,2)$ is fixed and that the lines $$\textbf{a}: y-z=y+z-2=0$$
and $$\textbf{b}: z-1=x-z=0$$ are invariant.
So, I think I've found successfully ONE affine transformation satisfying all of them at the same time, which is $$T(x,y,z)=(-x-4z+6,-y+2z,z)$$
I did it sending $(-1,2,2)$ to itself, sending the intersection point of both lines $(1,1,1)$ to itself and choosing two random points $(0,1,1)$ and $(1,0,1)$ from the lines and associating them to another random point $(2,1,1)$ and $(1,2,1)$ on their respective lines, respectively.
Of course, point $(-1,2,2)$ is fixed and I've checked several points contained in the lines above and seen that their images are still on the same lines.
The problem is that, supposedly, I think that there could be more than one affine transformation satisfying those conditions at the same time. Maybe, a family of affine transformations depending on a parameter. How can I get all possible affine maps satisfying the conditions given or prove that an affine map is unique under a certain set of circumstances (although this might not be the case in this exercise)?
| You’ve made a good start by identifying a second fixed point of the transformation. The required family of affine transformation has two degrees of freedom remaining. To see this, consider a nonsingular affine transformation on a line: it consists of a scaling followed by a translation. For each of the two invariant lines, then, if we choose an arbitrary nonzero scale factor, the translation part of the restricted affine transformation is determined by the fixed intersection point, so we can take those two arbitrary scale factors as parameters of the transformation.
I find it convenient to work in homogeneous coordinates for problems like these. An affine transformation can be represented by a single matrix, and its invariant flats can be described in terms of eigenvectors of this matrix. So, we seek a matrix of the form $$M = \begin{bmatrix}m_{11}&m_{12}&m_{13}&m_{14}\\m_{21}&m_{22}&m_{23}&m_{24}\\m_{31}&m_{32}&m_{33}&m_{34}\\0&0&0&1\end{bmatrix}.$$ The two fixed points are eigenvectors of $M$ with eigenvalue $1$. The directions of the two invariant lines must also be preserved, so the homogeneous vectors that represent their directions are also eigenvectors of $M$, but with arbitrary nonzero eigenvalues (these are the scale factors from above). The direction vectors of the two lines can be found by computing the cross products of the normals of the planes in their defining equations.
Using the “bulk” eigenvector equation $MX=X\Lambda$, we find that $$M = \begin{bmatrix}-1 & 1 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0\end{bmatrix} \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&\lambda&0\\0&0&0&\mu\end{bmatrix} \begin{bmatrix}-1 & 1 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0\end{bmatrix}^{-1} = \begin{bmatrix}\lambda & 0 & 2(\lambda-1) & 3(1-\lambda) \\ 0 & \mu & 1-\mu & 0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix},$$ i.e., $$T:\begin{bmatrix}x\\y\\z\end{bmatrix} \mapsto \begin{bmatrix} \lambda x+2(\lambda-1)z+3(1-\lambda) \\ \mu y+(1-\mu)z \\ z\end{bmatrix}$$ for $\lambda\mu\ne0$. The transformation you found is obtained by setting $\lambda=\mu=-1$.
By construction, the two fixed points are mapped to themselves, but it’s worth checking anyway to ensure that you haven’t made any algebraic errors along the way. In addition, $T(1,0,0)=(\lambda,0,0)$ and so $(1,1,1)+t(1,0,0)$ is mapped to $(1,1,1)+\lambda t(1,0,0)$, which is obviously another parameterization of the same line. A similar simple calculation verifies that the other line is also mapped onto itself.
You can also make a slight modification to your solution method to get the general solution: Instead of mapping $(0,1,1)$ and $(1,0,1)$ to specific points on the respective lines, map them to the arbitrary points $(1,1,1)+\lambda(1,0,0)$ and $(1,1,1)+\mu(0,1,0)$, respectively, and then carry out the same calculations that you made to obtain the one transformation that you did find.
| {
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Understanding the solution to this polynomial estimation problem Show that there exist $K, N > 0$ such that for $x ∈\mathbb{R}$,
$$x ≥ N \implies
\frac{3x^
2 − 4x + 8}{
5x + 6}
≥ Kx.$$
Solution:
For $x ≥ 4,$ we have $4x ≤ x^
2$
so that
$$3x^
2 − 4x + 8 ≥ 3x^
2 − x^
2 = 2x^
2.$$
Similarly, for $x ≥ 1,$ we have
$$5x + 6 ≤ 11x.$$
Therefore, when $x ≥ 4,$ we have
$$\frac{3x^
2 − 4x + 8}{
5x + 6}
≥\frac{
2x^
2}{
11x}
=
\frac{2}
{11} x,$$
and so we can take $N = 4$ and
$K = \frac{2}{
11} .$
I'm struggling to follow the logical steps in this solution. I'm not sure why the inequality of $x ≥ 4$ and $x ≥ 1$ has been chosen. Can someone please break down this solution and show where the steps come from?
| Essentially the method of attack that was used in this proof is to find a quadratic function that is less than the numerator and a linear function that is greater than the denominator. Then when you divide these you get a linear function which is less than the original $\frac{3x^2-4x+8}{5x+6}$.
In layman's terms the question is asking if it is possible to find some real number $N$ such that $\frac{3x^2-4x+8}{5x+6}$ is greater than some linear function $Kx$ for all $x$ greater than or equal to $N$.
So again we are looking for a quadratic equation which is less than $3x^2-4x+8$ for some values of $x > 0$. Notice that $3x^2 + 8 \geq 3x^2 $ for all $x > 0$. Now we need to account for the $-4x$ term still in the numerator. There's actually an infinite amount of choices we could do for the next step. In the proof you presented the author realized that $x^2$ would be greater than or equal to $4x$ eventually (namely at $x = 4$). So from the inequality above we note that $3x^2 - 4x + 8 \geq 3x^2 - x^2 = 2x^2$ when $x \geq 4$ (because we subtracted something greater from the RHS than the LHS).
Now for the next step again we have an infinite amount of choices, however the author realized that $5x +6 \leq 11x$ when $x \geq 1$. Of course they could just have easily said $5x+6 \leq 12x$ or an infinite number of other choices for that matter (thus finding a different value of $K$) but they chose this. Now since $5x + 6 \leq 11x$ when $x \geq 1, $ it is certainly also true when $x \geq4.$
Now for the final step, we can obtain the equation $\frac{3x^2 - 4x + 8}{5x+6} \geq \frac{2x^2}{11x} = \frac{2}{11}x$ for $x \geq 4$ because the numerator of the LHS is greater than the numerator of the RHS and the denominator of the LHS is less than the numerator of the RHS for $x\geq4$ as we showed above.
Thus we can pick $N = 4$ and $K = \frac{2}{11}$. To make sure you understand it try and find another value of $N$ and $K$, for instance by realizing that $3x^2 - 4x + 8 \geq 3x^2 -2x^2 = x^2$ for $x \geq 4$ in the first step.
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If irrational numbers are uncountable, then why did I find this? I understand that irrational numbers are uncountable. I've seen the proof and it makes perfect sense. However, I came up with this (most likely false) proof that says that they're countable. Chances are, I made a mistake and the proof doesn't mean anything, but I just want to be sure. Here is the proof:
You can't find a solid chunk (range of numbers) that does not contain a rational number. Which means that irrational numbers are just points on the number line, not lines. Which means you just need to name all the points to count the irrationals.
What is the mistake here?
| "You can't find a solid chunk (range of numbers) that does not contain a rational number." - True, this is just the statements that the rationals are dense in the reals (and also irrationals).
"Which means that irrational numbers are just points on the number line, not lines." - True
"Which means you just need to name all the points to count the irrationals." - Basically saying "we need to give a name to everything in this bucket". As @Deepak noted, this is the hard part. Note that whatever (finite!) "name" you come up with, it can be mapped to the naturals via eg dictionary sorting. This is basically the statement that the computable irrationals are still countable - whether "sqrt of pi" or "8th root of polynomial: ___". Beyond these are an uncountable sea of irrationals that cannot be computed/described/etc by any finite means. How do you plan to name these?
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100 numbers chosen on unit interval 100 numbers are chosen from the interval [0, 1], independently of each other. What is the probability that the 2nd largest # is <1/2? My reasoning was a) all 100 chosen are less then 1/2
b) 99 are less then 1/2 and one is more
This would give (1/2)^100 OR the probability of 99 less then 1/2 and one greater . Would i not need binomial theorem since im also factoring the number if ways to CHOOSE the 99 numbers ??
| I would go about it using order statistics. Let the 100 independent random variables are $X_1,\,X_2,\,\cdots,\,X_{100}$ where all are uniformly distributed in $[0,\,1]$. Arrange them as $$X_{(1)}\leq X_{(2)}\leq\cdots\leq X_{(99)}\leq X_{(100)}$$ Then you need to find $$Pr\left[X_{(99)}<\frac{1}{2}\right]=\sum_{i=99}^{100}{100\choose i}F_X^i(0.5)\left[1-F_X(0.5)\right]^{100-i}$$ where $F_X(x)$ is the CDF of the random variables.
| {
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Why SVD is not unique but the Moore-Penrose pseudo inverse is unique? I feel confused about the uniqueness of the Moore-Penrose inverse generated from SVD.
For any matrix $A$, if $X$ satisfied $$AXA=A, XAX=X, (AX)^\mathrm{T}=AX, (XA)^\mathrm{T}=XA $$then $X$ is called the Moore-Penrose inverse of $A$.
If $A$ has the SVD(singular value decomposition)$$A=P\left[\begin{matrix}\Lambda_r&0\\0&0\end{matrix}\right]Q^\mathrm{T}$$
then it is easy to prove that$$A^+ = Q\left[\begin{matrix}\Lambda_r^{-1}&0\\0&0\end{matrix}\right]P^\mathrm{T}$$ is a Moore-Penrose inverse.
If $X$ and $Y$ are both Moore-Penrose inverse of $A$, from the equation$$X=XAX=X(AX)^\mathrm{T}=XX^\mathrm{T}A^\mathrm{T}=XX^\mathrm{T}(AYA)^\mathrm{T}=X(AX)^\mathrm{T}(AY)^\mathrm{T}=(XAX)AY=XAY=(XA)^\mathrm{T}YAY=A^\mathrm{T}X^\mathrm{T}A^\mathrm{T}Y^\mathrm{T}Y=A^\mathrm{T}Y^\mathrm{T}Y=(YA)^\mathrm{T}Y=YAY=Y$$
we can see that the Moore-Penrose inverse is unique.
However, the Moore-Penrose inverse depends on the SVD and SVD is not unique. How to explain it?
| The non-uniqueness of SVD can be characterized as follows: suppose that $A = P_0 \Sigma Q_0^T$ is one SVD of $A$. Moreover, suppose that the singular values of $A$ are $s_1$ with multiplicity $k_1$, $s_2$ with multiplicity $k_2$, and so forth, with $s_m = 0$ having multiplicity $k_m = n - r$. That is, we have
$$
\Lambda_r = \pmatrix{s_1 I_{k_1} \\ & \ddots \\ && s_{m-1} I_{k_{m-1}}}, \quad \Sigma = \pmatrix{\Lambda_r \\ & 0_{k_m}}
$$
Then $A = P\Sigma Q^T$ will be a singular value decomposition of $A$ if and only if there exists an orthogonal matrix $U$ such that $P = P_0 U, Q = Q_0U$, and $U$ is a block-diagonal orthogonal matrix of the form
$$
U = \pmatrix{U^{(1)}\\ & \ddots \\ && U^{(m)}}
$$
where $U^{(j)}$ is (orthogonal and) of size $k_j \times k_j$.
With that in mind: if you'd like to prove that the pseudoinverse as constructed from SVD is well-defined (that is, uniquely defined regardless of one's choice of SVD), then it suffices to show that for any choice of $U$ of the form prescribed above, we have
$$
[Q_0U] \pmatrix{\Lambda_r^{-1} \\ & 0} [P_0U]^T = Q_0 \pmatrix{\Lambda_r^{-1} \\ & 0} P_0
$$
It is straightforward (but in my opinion tedious) to show that this holds if we use the block-structure of $\Lambda_r^{-1}$ and block-matrix multiplication.
| {
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Why we can represent automorphisms in $\text{Gal}(\Bbb Q(\sqrt[4]{2},i)/\Bbb Q)$ as permutations in $S_4$ but not $S_8$? The splitting field of $x^4-2$ over $\Bbb Q$ is $G=\text{Gal}(\Bbb Q(\sqrt[4]{2},i)/\Bbb Q)$. By primitive element theorem, $K=\Bbb Q(\alpha)$ for some $\alpha$ and $[K:\Bbb Q]=8$. So I know that the minimal polynomial of $\alpha$ over $\Bbb Q$ has degree $8$. And at the same time there are eight automorphisms in $G$.
However, one thing I don't understand is that some authors labeled the roots $\sqrt[4]{2},~\sqrt[4]{2}i,~-\sqrt[4]{2},~-\sqrt[4]{2}i$ of $x^4-2$ as $1,~2,~3,~4$, and say that every automorphisms in $G$ can be represented as a permutation. For example, $\sigma_1=(1~~2~~3~~4)$. How can we do this? Why we can only analyze how the roots $\sqrt[4]{2},~\sqrt[4]{2}i,~-\sqrt[4]{2},~-\sqrt[4]{2}i$ is affected by the automorphism, and then we can decide it? Maybe there are different $\sigma,~\sigma'\in G$ such that $\forall r\in\{\sqrt[4]{2},~\sqrt[4]{2}i,~-\sqrt[4]{2},~-\sqrt[4]{2}i\},~\sigma(r)=\sigma'(r)$.
Picture:
| Remember that if $F$ is the ground field, and $p(x)$ is an irreducible polynomial over $F$ of degree $n\geq 1$ without repeated roots, let $K$ be the splitting field of $p(x)$. $K/F$ is Galois. You have that the Galois group of $K/F$, say $G$, it is embedded on $S_n$, this is because $G$ permutes the roots.
In your problem, $F=\mathbb{Q}$ and $K$ is the splitting field of $p(x)=x^4-2$, since its roots are $\sqrt[4]{2}$, $-\sqrt[4]{2}$, $i\sqrt[4]{2}$, $-i\sqrt[4]{2}$.
For a better understanding. The Galois group acts transitively on the roots of that polynomial (it is important the irreducibility). If you are not clear about the Galois group yet, you can take intermediate fields, study the irredubile polynomial on the tower of fields and you will know that the Galois Group in some extension of the tower of fields acts transitively on the root and fix the elements on the base field. For example, in this case, if you want to use that method, consider the tower of fields $\mathbb{Q}\left(\sqrt[4]{2},i\right)\supset\mathbb{Q}\left(\sqrt[4]{2}\right)\supset \mathbb{Q}$, since $\mathbb{Q}\left(\sqrt[4]{2},i\right)/\mathbb{Q}\left(\sqrt[4]{2}\right)$ is Galois (splitting field of $x^2+1$), you know now that theres is a $\sigma$ in $\mbox{Gal}\left(\mathbb{Q}\left(\sqrt[4]{2},i\right)/\mathbb{Q}\left(\sqrt[4]{2}\right)\right)$ such that send $i\mapsto -i$ (acts on the roots of $x^2+1$) and leave $\mathbb{Q}\left(\sqrt[4]{2}\right)$ fixed, but the Galois group of that extension if a subgroup of the Galois group of $K/\mathbb{Q}$, so $\sigma$ lie to $\mbox{Gal}\left(K/\mathbb{Q}\right)$
| {
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Is this map chain-transitive Let $f:\mathbb R^2→\mathbb R^2$ be a continuous self-map and let $δ$ be a positive real number. A (finite or infinite) sequence $(x_{n})_{n≥0}$ is a $δ$-chain if $$d(f(x_{n}),x_{n+1})<δ$$ for all $n$.
The map $f$ is called chain-transitive if for every $x,y∈\mathbb R^2$ and every $δ>0$ there is a finite $δ$-chain $x₀,...,x_{n}$ such that $x₀=x$ and $x_{n}=y$.
Assuming a map g verify: $g^{k}(x)$ tends to infinity as $k→∞$ for all $x \in \mathbb R^2$, i.e., the successive iterations of $x$ by $g$ goes to infinity, or the orbit of the points $x \in \mathbb R^2$ diverge.
My question is: is this map $g$ chain-transitive?
| Such a $g$ might not be Chain transitive.
Counter-example: $g(a, b) = (a+1, b)$. Then there does not exists a $\delta$ transitive chain between $x = (0,0)$ and $y = (1,0)$ whenever $\delta <1/2$: No matter what one choose $x_n = (a_n, b_n)$, one has $a_n \ge n/2$.
| {
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How to show that $\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx =\frac{\pi}{8}$ I was reading Advanced Integration Techniques, and found that$$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx =\frac{\pi}{8}$$
The book provides one method using residue theorem and Laurent expansion. However, I wonder if there are other techniques that I can use to evaluate this integral.
The most direct method would be solving the integral and plug in the limits.
$$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx = \left[\frac{\arcsin(2x-1)+\sqrt{(1-x)x}(4x-2)}{8}\right]_{0}^{1}=\frac{\pi}{8}$$
| J.G. has the elementary method down. If you see a $1-x^2$ term inside your integrand, it might be wise to give a trig substitution a try. In this case, letting $x=\sin^2\theta$ works out beautifully.
There’s another less elementary way by utilizing the beta function and the Gamma function. Let me know if you need a proof (I proudly have an elementary proof of it).
Beta Function:$$\operatorname{B}\left(m,n\right)=\int\limits_0^1\mathrm dx\,x^{m-1}(1-x)^{n-1}=\frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$
The integral under question is then simply$$\begin{align*}\mathfrak{I} & =\operatorname{B}\left(\frac 32,\frac 32\right)\\ & =\frac {1}{2}\left(\frac {\sqrt{\pi}}2\right)^2\end{align*}$$
Thus$$\int\limits_0^1\mathrm dx\, \sqrt{x(1-x)}\color{blue}{=\frac {\pi}8}$$
| {
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True or False, diagonalization problem
Let $B_c= \left\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\right\}$, $T:\mathbb{R}^4\rightarrow \mathbb{R}^4$ a linear operator such that \begin{equation} \det\left[T-I\lambda\right]_{B_{c}}=(2-\lambda)^4\qquad \text{and}\qquad T((0,0,0,1)) = (1,0,0,1)\end{equation}
Is T diagonalizable?
I've tried to use the following reasoning: Since the algebraic multiplicity of the eigenvalue $2$ is $4$, we know that $T$ is diagonalizable if (and only if) the dimension of the eigenspace associated with the eigenvalue $2$ is $4$. But we also know that there's a vector in $\mathbb{R}^4$ that don't belong to $S_{2}$ (because $(0,0,0,1)$ is not an eigenvector). Can I use this argument to show that $T$ is not diagonalizable? Thanks in advance
| Since $T((0,0,0,1)) \neq 2 \cdot (0,0,0,1)$ we know that $(0,0,0,1)$ isn't in the eigenspace $\operatorname{Ker}(T-2I)$. The dimension of the eigenspace is therefore strictly less than the dimension of $\mathbb{R}^4$.
| {
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find the Jordan basis of a matrix I'm trying to find the Jordan basis of the matrix $$A =\begin{bmatrix} 8 & 1 & 2 \\ -3 & 4 & -2\\ -3 & -1 & 3\end{bmatrix}$$ I've got the characteristic equation to be $CA(x) = (5-x)^3$ and hence the eigenvalue to be $5$. I started by finding $v_1$ such that $(A-5I)v_1=0$ and chose $v_1 = \begin{bmatrix} 1 \\ -3 \\ 0\end{bmatrix}$. I then tried to find a $v_2$ such that $(A-5I)^2v_2 = 0$ but $(A-5I)^2$ is the zero matrix so can $v_2$ be any vector in $\mathbb{R}^3$?
More generally, when trying to find a Jordan basis for a matrix $A$, what do you do when $(A-\lambda I)^i=0$ for some $i$? Also what do you do when there is no solution to $(A-\lambda I)^iv_i = 0$ for a particular $i$?
| The eigenspace $E_5$ has dimension $2$ (so the Jordan form of the matrix will have $2$ Jordan blocks) since
$$A-5I=\begin{bmatrix}
3&1&2\\-3&-1&-2\\-3&-1&-2\end{bmatrix}
$$
and it is defined by the single equation $\;3x+y+2z=0$.
You should attack the problem backwards: begin with choosing a vector $v_3=(x,y,z)$,
in $\ker(A-5I)^2\smallsetminus E_5=\mathbf R^3\smallsetminus E_5$, i.e. such that
$$3x+y+2z\ne 0,\enspace \text{say }\enspace v_3=(1,0,-1),$$
and set $\;(A-5I)v_3=v_2=(1,-1,-1)$ ; $v_2$ belongs to the eigenspace $E_5$.
Last you have to complete $v_2$ with another vector $v_1$ in the eigenspace which is linearly independent from $v_2$. The vector you've found – $v_1=(1,-3,0)$ is fine. In the basis $\mathcal B=(v_1,v_2,v_3)$ the matrix of $A$ becomes by construction:
$$J=\begin{bmatrix}
5&0&0\\0&5&1\\0&0&5\end{bmatrix}. $$
| {
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Prove that $ \binom nm \sum_{i=0}^{m} (-1)^i\frac{\binom mi}{n-m+1+i}=\frac{1}{1+n}$ I want to prove that
$$
\binom nm \sum_{i=0}^{m} (-1)^i\frac{\binom mi}{n-m+1+i}=\frac{1}{1+n}\\
$$
$$
\binom nm [ \frac{\binom m0}{n-m+1} +(-1)^1\frac{\binom m1}{n-m+2}+...
..+(-1)^m\frac{\binom mm}{n+1}]
=\frac{1}{1+n}\\
$$
*
*Assume that $m=1$,then it's equal to $1/(n+1)$.
$$
\binom n1 [ \frac{1}{n} -\frac{1}{n+1}] =\frac{1}{1+n}\\
$$
*Assume that $m=2$,then it's equal to $1/(n+1)$.
$$
\binom n2 [ \frac{1}{n-1} -\frac{2}{n}+\frac{1}{n+1}] =\frac{1}{1+n}\\
$$
How to prove that for $\forall m$, this formula is equal to $1/(n+1)$？
| Hint. Note that
$$x^{n-m}(1-x)^m=\sum_{i=0}^{m} (-1)^i \binom mi x^{n-m+i}$$
Now integrate with respect to $x$ over the interval $[0,1]$ and recall the definition of Beta function.
| {
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Do all projections matrices take this form? Do all projection matrices take the form $P = A{(A^TA)}^{-1}A^T$? If so, can you help me derive it and explain it intuitively?
| $A(A^TA)^{-1}A^T$ is symmetric, but not all projection matrices are symmetric --- such as $\pmatrix{1&1\\ 0&0}$. Thus the answer is clearly no.
It is true, however, that all orthogonal (with respect to the usual inner product) projection matrices over $\mathbb R$ can be written in the form of $A(A^TA)^{-1}A^T$. By definition, if $P\in M_n(\mathbb R)$ is an orthogonal projection, then $P|_U=\operatorname{id}$ and $P|_{U^\perp}=0$ for some subspace $U\subseteq\mathbb R^n$. Let $A$ be any matrix whose columns form a basis of $U$ (any basis will do; it doesn't have to be orthonormal). Then $A^TA$ is nonsingular and $A(A^TA)^{-1}A^Tv=0$ for every $v\in U^\perp$. Also, since the columns of $A$ span $U$, every vector $u\in U$ can be written as $Ax$ for some $x\in\mathbb R^n$. Therefore
$$
\left(A(A^TA)^{-1}A^T\right)u=\left(A(A^TA)^{-1}A^T\right)(Ax)=\left(A(A^TA)^{-1}A^TA\right)x=Ax=u
$$
for every $u=Ax\in U$. Hence $P$ and $A(A^TA)^{-1}A^T$ agree everywhere on $\mathbb R^n$, i.e. $P=A(A^TA)^{-1}A^T$.
| {
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Product of two polytopes is a polytope Please have a look at my attempt for this problem.
Let $x = \begin{pmatrix} x_1\\ x_2 \\ \end{pmatrix}, x_1 \in P_1, x_2 \in P_2$.
I want to show that $x \in conv\{P_1 \times P_2\}$, i.e. $x$ can be represented as the convex combination of some points of $P_1 \times P_2$.
Without loss of generality, suppose that $d_1 \geq d_2$. By Caratheodory's theorem, $x_1$ can be represented by at most $d_1 + 1$ points of $P_1$, i.e. there exists $\{v_1,...,v_{d_1+1}\} \subset P_1$, and $\alpha_1,...,\alpha_{d_1+1}$; $ \alpha_i \geq 0$, $ \sum \alpha_i=1$, such that:
$$x_1 = \alpha_1v_1+...+\alpha_{d_1+1}v_{d_1+1}$$
$x_2$ can be represented similarly, with points $\{w_1,...,w_{d_2+1},...,w_{d_1+1}\} \subset P_2$, and coefficients $\beta_i$'s, such that $\beta_{d_2+2},...,\beta_{d_1+1}$(if exist) are all $0$'s:
$$x_2 = \beta_1w_1+...+\beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1}$$
So now we have $x$ as:
$$x = \begin{pmatrix} x_1\\ x_2 \\ \end{pmatrix} = \begin{pmatrix} \alpha_1v_1+...+\alpha_{d_2+1}v_{d_2+1}+\alpha_{d_2+2}v_{d_2+2}+...+\alpha_{d_1+1}v_{d_1+1}\\ \beta_1w_1+...+\beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1} \\ \end{pmatrix}$$
It is here that I got stuck. The points $\begin{pmatrix} v_i\\ w_i \\ \end{pmatrix}$ above are certainly in $P_1 \times P_2$, but are the coefficients right? Shouldn't the coefficients be in $\mathbb{R}$, instead of $\mathbb{R}^2$?
===================================
Edit: the polytopes here are all convex polytopes.
Edit 2: Caratheodory's theorem: https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(convex_hull)
| There is a much easier way to do that.
Definition
let $v,w\in \Bbb R^n$. We define $$v\succeq w\iff v_i\ge
w_i\quad,\quad 1\le i\le n$$
According to this definition, a closed half-space can be defined as following$$\{x|a^Tx\le b\}$$when $a,x\in\Bbb R^n$ and $b\in \Bbb R$. Therefore an intersection of finite number of half-spaces would become $$\{x\ \ |\ \ A^Tx\preceq b\}=\{x\ \ |\ \ a_i^Tx\preceq b_i \ \ ,\ \ 1\le i\le m\}$$when $x\in\Bbb R^n , A\in \Bbb R^{m\times n},b\in\Bbb R^m$ and $a_i^T$s are the rows of $A$. Then a convex polytope can be easily defined as$$P=\{x\ \ |\ \ A^Tx\preceq b\}$$Hence this problem, let $$P_1=\{x\ \ |\ \ A_1^Tx\preceq b_1\}\\P_2=\{y\ \ |\ \ A_2^Ty\preceq b_2\}$$where $${x\in\Bbb R^n,y\in\Bbb R^m \\ A_1\in \Bbb R^{m_1\times n},A_2\in \Bbb R^{m_2\times m}\\ b_1\in\Bbb R^{m_1},b_2\in\Bbb R^{m_2}}$$Now let $$P_3=P_1\times P_2{=\Big\{\begin{bmatrix}x\\y\end{bmatrix}\ \ \Big|\ \ x\in P_1\ \ ,\ \ y\in P_2\Big\}\\=\Big\{\begin{bmatrix}x\\y\end{bmatrix}\ \ \Big|\ \ A_1x\preceq b_1\ \ ,\ \ A_2y\preceq b_2\Big\}\\=\Big\{\begin{bmatrix}x\\y\end{bmatrix}\ \ \Big|\ \ \begin{bmatrix}A_1&0\\0&A_2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\preceq \begin{bmatrix}b_1\\b_2\end{bmatrix}\Big\}\\=\Big\{\begin{bmatrix}x\\y\end{bmatrix}\ \ \Big|\ \ A_3\begin{bmatrix}x\\y\end{bmatrix}\preceq b_3\Big\}\\=\Big\{z\in \Bbb R^{m+n}\ \ \Big|\ \ A_3z\preceq b_3\Big\}}$$where $$A_3=\begin{bmatrix}A_1^{m_1\times n}&0^{m_1\times m}\\\\0^{m_2\times n}&A_2^{m_2\times m}\end{bmatrix}_{(m_1+m_2)\times(m+n)}$$and $$b_3=\begin{bmatrix}b_1^{m_1\times 1}\\b_2^{m_2\times 1}\end{bmatrix}_{(m_1+m_2)\times1}$$with the coordinates $$z=\begin{bmatrix}x_{n\times 1}\\y_{m\times 1}\end{bmatrix}_{(m+n)\times1}$$
Since we could express $P_3$ as $\Big\{z\in \Bbb R^{m+n}\ \ \Big|\ \ A_3z\preceq b_3\Big\}$ with some matrices $A_3$ and $b_3$, then $P_3$ is also a convex polytope and we conclude that
If $P_1$ and $P_2$ are convex polytopes of dimensions $m$ and $n$ respectively, then their product $P_1\times P_2=\Big\{\begin{bmatrix}x\\y\end{bmatrix}\ \ \Big| \ \ x\in P_1 \ \ , \ \ y\in P_2\Big\}$ is a convex polytope of dimension $m+n$.
| {
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Explanation of Markov transition function Here the definition of my course (in the picture below). Could someone explain me the Chapman-Kolomogorov equation ? I don't really understand what it mean. Also, I tried to make a parallel with discrete Markov chain, I don't see the link between continuous and discrete Markov chain. Is the motivation behind the same ? (in discrete time $(X_n)$ is a Markov chain if $$\mathbb P\{X_{n+1}=x\mid \sigma (X_{k}\mid k\leq n)\}=\mathbb P\{X_{n+1}=x\mid X_n\}.$$ Also, $P_{s,t}(x,dy)$ is the regular version of the conditional distribution of $X_t$ given $X_s$... I'm not really sure what it mean, is it $$P_{s,t}(x,dy)=\mathbb P\{X_t\mid X_s\} \ \ ?$$
But it doesn't really make sense.
| You can think of a continuous-time Markov process as being fully characterized by both an "embedded" discrete-time Markov chain that governs the probabilities of transitions between states (call them $Q_{ij}$), as well as some holding time parameters $\lambda_i$ that represent the average rates at which one transitions out of a state $i$. (Those holding times are always distributed exponentially, so knowing $\lambda_i$ is enough to characterize their distribution.)
You can then construct an evolution equation, sometimes called a master equation, for the continuous-time transition probabilities $P_{ij}(t)$ using these parameters.
The Chapman-Kolmogorov equations for continuous-time Markov processes are "the same thing" they were in the discrete version: an identity for the transition probabilities based on conditioning on an intermediate step and exploiting the Markov property.
| {
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Are uniformly equivalent metrics with the same bounded sets Holder equivalent? This is a follow-up to my question here. Let $d_1$ and $d_2$ be two metrics on the same set $X$. Then $d_1$ and $d_2$ are uniformly equivalent if the identity maps $i:(M,d_1)\rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)\rightarrow(M,d_1)$ are uniformly continuous. And $d_1$ and $d_2$ are Holder equivalent if there exist constants $\alpha\in (0,1]$ and $C,D>0$ such that $C(d_1(x,y))^\alpha\leq d_2(x,y)\leq D (d_1(x,y))^\alpha$ for all $x,y\in X$.
Now if $d_1$ and $d_2$ are Holder equivalent, then they are uniformly equivalent and they have the same bounded sets. My question is, is the converse true? That is, if $d_1$ and $d_2$ are uniformly equivalent and have the same bounded sets, then are they Holder equivalent?
If not, is there an example of metrics which are uniformly equivalent and have the same bounded sets but are not Holder equivalent?
| Here is a counterexample. Let $X = \mathbb{R}$ and $d_1(x,y) = \lvert y - x \rvert$. Define $d_2(x,y) = \lvert y - x \rvert$ if $\lvert y - x \rvert \le 1$ and $d_2(x,y) = \sqrt{\lvert y - x \rvert}$ if $\lvert y - x \rvert \ge 1$.
An obvious property of $d_2$ is that $\lvert y - x \rvert \le \lvert y' - x' \rvert$ implies $d_2(x,y) \le d_2(x',y')$. We shall moreover need the following well-known fact:
$(*)$ If $1 \le a \le b$, then $\sqrt{b} - \sqrt{a} \le b - a$.
Let us verify that $d_2$ is a metric. The only thing which is not trivial is the triangle inequality. For $x,y,z \in \mathbb{R}$ we have to show $d_2(x,y) \le d_2(x,z) + d_2(z,y)$. This is trivially true if $y = x$. In case $y \ne x$ we may w.l.o.g. assume that $y > x$, i.e. $y = x + r$ with $r > 0$.
(a) $z \le x$. Then $\lvert y - x \rvert \le \lvert z - x \rvert$, hence $d_2(x,y) \le d_2(x,z)$. The triangle inequality follows.
(b) $z \ge y$. Similar!
(c) $x < z < y$.
(c1) $r \le 1$. This reduces to the triangle inequality for $d_1$.
(c2) $r > 1$. Write $z = x + s$ with $0 < s < r$.
(c2.1) $s \le 1$ and $r - s \le 1$. Then we have to show $\sqrt{r} \le s + (r - s) = r$ which is true since $r > 1$.
(c2.2) $s \le 1$ and $r - s > 1$. Then we have to show $\sqrt{r} \le s + \sqrt{r - s}$. This follows from $(*)$.
(c2.3) $s > 1$ and $r - s \le 1$. Then we have to show $\sqrt{r} \le \sqrt{s} + (r - s) $. This follows agaim from $(*)$.
(c2.4) $s > 1$ and $r - s > 1$. Then we have to show $\sqrt{r} \le \sqrt{s} + \sqrt{r - s}$. This is obvious (take the squares of both sides).
$d_1$ are $d_2$ are uniformly equivalent because $d_2(x,y) \le d_1(x,y)$ and $d_2(x,y) < \min(1,\epsilon)$ implies $d_1(x,y) < \epsilon$ (in fact, we have $d_2(x,y) < 1$ which is only possible when $\lvert y - x \rvert < 1$ so that $d_1(x,y) = d_2(x,y) < \epsilon $).
$d_1$ and $d_2$ have the same bounded sets. If $B$ is bounded with respect to $d_1$, then it is obviously respect to $d_2$. Let $B$ be unbounded with respect to $d_1$. Then there exist $x_n,y_n \in B$ such that $d_1(x_n,y_n) \ge n$. But then $d_2(x_n,y_n) \ge \sqrt{n}$, hence $B$ be unbounded with respect to $d_2$.
$d_1$ and $d_2$ are not Hölder equivalent. Assume there are $C > 0$ and $\alpha \in (0,1]$ such that $d_1(x,y) \le C (d_2(x,y))^\alpha$. For $\lvert y - x \rvert \ge 1$ this means $\lvert y - x \rvert \le C \lvert y - x \rvert^{\alpha/2}$, i.e. $\lvert y - x \rvert^{1 - \alpha/2} \le C$. But $1 - \alpha/2 \in [1/2,1)$, hence $\lvert y - x \rvert^{1 - \alpha/2}$ cannot bounded by any constant $C$. This is a contradiction.
| {
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"url": "https://math.stackexchange.com/questions/3058784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Angular velocity as eigenvector of rotation map In physics the angular velocity $\omega$ usually is defined as follows:
For a map $B: \mathbb{R} \to SO(3)$ define $\Omega(t):=\dot{B}(t)B^t(t)$. This is a scew-symmetric matrix and hence there is a vector $\omega(t) \in \mathbb{R}^3$ satisfying $\Omega(t) x=\omega(t) \times x$ for $x\in\mathbb{R}^3$.
Does $\omega$ always point in the direction of the "momentary axis of rotation of $B$"? That is, is $\omega(t)$ an eigenvector of $B(t)$? I've tried to prove this by writing $\omega_j(t) = \frac{1}{2}\sum_{k,l=1}^3\epsilon_{jkl}\Omega_{kl}(t)$ (were $\epsilon_{jkl}$ is the Levi-Cita symbol). Then $$(B(t)\omega(t))_j=-\frac{1}{2}\sum_{k,l,m=1}^3\epsilon_{ikl}B_{ji}(t)\epsilon_{ikl}\Omega_{kl}(t)=-\frac{1}{2}\sum_{k,l,m,n=1}^3\epsilon_{ikl}B_{ji}(t)\dot{B}_{kn}(t)B_{ln}(t).$$ I don't know how to proceed further. Is this even true?
| No this is not true in general. Take for example
$$ B = \begin{pmatrix} 0 & -\sin\phi & -\cos\phi\\ 0 & \cos\phi & -\sin\phi\\ 1 & 0 &0 \end{pmatrix}$$
Then
$$\dot B = \dot\phi\begin{pmatrix} 0 & -\cos\phi & \sin\phi \\ 0 & -\sin\phi & -\cos\phi \\ 1 & 0 & 0\end{pmatrix}
\quad\text{and}\quad \Omega = \dot\phi\begin{pmatrix} 0 & -1& 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \end{pmatrix}, \omega = \begin{pmatrix}0\\0\\-\dot\phi\end{pmatrix} $$
However, if $\dot\phi(0) \neq 0$ then $\omega(0)$ is not an eigenvector of $B(0)$.
There also is no reason why this should be true. The matrix $B$ represents an orientation, not a rotation. It's Eigenvectors are therefore meaningless. The "momentary rotation" of the object if given by $\exp(\Omega)$.
And indeed, since $\Omega \omega = \omega \times \omega = 0$ we have $\exp(\Omega)\omega = \omega$.
| {
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Suppose $\sum_{n=1}^{\infty}\sqrt{{a_{n}}/{n}}$ is convergent. Prove that $\sum_{n=1}^{\infty}a_{n}$ is also convergent. Let $\{a_{n}\}$ be a decreasing sequence of non-negative real numbers. Suppose $\sum_{n=1}^{\infty}\sqrt{\frac{a_{n}}{n}}$ is convergent. Prove that $\sum_{n=1}^{\infty}a_{n}$ is also convergent.
My thought is to use the direct comparison test but I think I'm struggling with showing that $a_{n}\leq\sqrt{\frac{a_{n}}{n}}$ $\forall$ n $\in\mathbb{N}$.
Any help would be great. Thank you!
| Let $b_n=\sqrt{a_n}$. Then $\{b_n\}_{n\geq 1}$ is a decreasing sequence of positive real numbers and we want to show that
$$ \sum_{n\geq 1}\frac{b_n}{\sqrt{n}}<+\infty\quad\Longrightarrow\quad \sum_{n\geq 1}b_n^2 < +\infty.$$
By the Cauchy-Schwarz inequality disguised as Titu's lemma we have
$$ \sum_{n=N+1}^{2N}\frac{b_n}{\sqrt{n}} \geq \frac{\left(\sqrt{b_{N+1}}+\ldots+\sqrt{b_{2N}}\right)^2}{\sqrt{N+1}+\ldots+\sqrt{2N}}\geq \frac{N^2 b_{2N}}{\frac{2}{3}(2\sqrt{2}-1)N\sqrt{N}}\geq \frac{4}{5}\sqrt{N}\,b_{2N} $$
hence
$$ \sum_{k\geq 0}2^{k/2} b_{2^k} $$
is convergent and $b_{2^k}=o(2^{-k/2})$. On the other hand
$$\begin{eqnarray*} \sum_{n=N+1}^{2N}b_n^2 &=& \sum_{n=N+1}^{2N}\sqrt{n}b_n\cdot\frac{b_n}{\sqrt{n}}\\&=&\sqrt{2N} b_{2N}\sum_{n=N+1}^{2N}\frac{b_n}{\sqrt{n}}+\sum_{n=N+1}^{2N-1}\left(\frac{b_n}{\sqrt{n}}-\frac{b_{n+1}}{\sqrt{n+1}}\right)\sum_{m=N+1}^{n}\frac{b_n}{\sqrt{n}}\end{eqnarray*} $$
by summation by parts, and assuming that $b_n$ is normalized in such a way that $\sum_{n\geq 1}\frac{b_n}{\sqrt{n}}=1$, the RHS is bounded by
$$\begin{eqnarray*}&&\sum_{n=N+1}^{2N-1}\frac{b_n}{\sqrt{n}}\left[\sqrt{2N} b_{2N}+\sum_{n=N+1}^{2N-1}\left(\frac{b_n}{\sqrt{n}}-\frac{b_{n+1}}{\sqrt{n+1}}\right)\right]\\&=&\sum_{n=N+1}^{2N-1}\frac{b_n}{\sqrt{n}}\left[\sqrt{2N} b_{2N}+\frac{b_{N+1}}{\sqrt{N+1}}-\frac{b_{2N}}{\sqrt{2N}}\right]\\&\leq& \sum_{n=N+1}^{2N-1}\frac{b_n}{\sqrt{n}}\left[\sqrt{2N} b_{2N}+\frac{5\sqrt{2}}{4N}\right]\\ &\leq&\frac{5\sqrt{2}}{4}\cdot\frac{N+2}{N+1}\sum_{n=N+1}^{2N}\frac{b_n}{\sqrt{n}}.\end{eqnarray*} $$
This detour gives a quantitative improvement of the other proofs: by summing on $N=2^k$ with $k\in\mathbb{N}$, then getting rid of the normalization assumption, we get
$$\boxed{\sum_{n\geq 1}b_n^2 \leq b_1^2+\color{blue}{\frac{15\sqrt{2}}{8}}\left(\sum_{n\geq 1}\frac{b_n}{\sqrt{n}}\right)^2} $$
and we may start wondering about the optimal constant that can replace $\frac{15\sqrt{2}}{8}$, like in Hardy's inequality. In this regard it makes sense to replace the short sums $\sum_{n=N+1}^{2N}$ with $\sum_{n=N+1}^{AN}$ and optimize on $A$. By considering $b_n=\frac{1}{n^{1/2+\varepsilon}}$, such that $\sum_{n\geq 1}\frac{b_n}{\sqrt{n}}$ is just barely convergent, we have that the blue constant cannot be improved beyond $\frac{7}{74}$.
Alternative approach: if a sequence $\{a_n\}_{n\geq 1}$ is such that $\sum_{n\geq 1}\lambda_n a_n$ is finite for any $\{\lambda_n\}_{n\geq 1}\in\ell^2$, then $\{a_n\}_{n\geq 1}\in\ell^2$. This is a consequence of the Banach-Steinhaus theorem, which can be proved independently by summation by parts (see page 150 of my notes). The constraints $0<a_{n+1}<a_n$ and $\sum_{n\geq 1}\frac{a_n}{\sqrt{n}}=C<+\infty$ should be more than enough to ensure that $\sum_{n\geq 1}\lambda_n a_n$ is finite for any $\Lambda\in\ell^2$, since $\sum_{n\geq 1}\frac{1}{n}$ is divergent.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inverting without actual inverse? I'm reading Ash's "Basic Abstract Algebra" and been trying to understand the following:
For which I guess the problem lies when we try to take an element $hk\in HK$, it's inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}\in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
My problem is the following, taking the definition of subgroup, we have:
*
*If $a,b\in H$, then $ab^{-1}\in H$.
Then, all the elements $k\cdot1=k$ and $1\cdot h=h$ are in $HK$, that is: We have all the elements of both $H,K$ in $HK$. By the definition, suppose we have $hk\in HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
It's a bit odd because it seems we can actually invert $hk$ by performing the two operations I mentioned before but $(hk)^{-1}$ is not actually in $HK$.
| $\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Allow me first of all to question your statement
when we try to take an element $hk\in HK$, its inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}\in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
This is not true. Take for instance $G = S_{3}$, $H = \Span{(1 2)}$, $K = \Span{(1 2 3)}$, $h = (1 2)$, $k = (1 2 3)$. Then $h^{-1} k^{-1} = (2 3) \ne (1 3) = k^{-1} h^{-1} \in H K = S_{3}$.
As to your second statement
suppose we have $hk\in HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
I fail to see the point. Sure, $h k$ has an inverse $k^{-1} h^{-1}$ in $G$, and that's what you are checking. But to show that $H K$ is a subgroup you will have to show that (under suitable conditions) this inverse lies precisely in $H K$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $\left|\sum_{k=0}^{n-1}\sin(2k+1)x\right|$ is bounded I'm currently studying Fourier series using a textbook which unfortunately provides only partial solutions to its explanations. To show the uniform convergence of a Fourier series I need to prove the assumption from the title above. Any help is very much appreciated.
| You can calculate this sum explicitly:
\begin{align}
\sum_{k=0}^{n-1}\sin(2k+1)x &= \operatorname{Im}\left(\sum_{k=0}^{n-1}e^{(2k+1)ix}\right)\\
&= \operatorname{Im}\left(e^{ix}\sum_{k=0}^{n-1}\left(e^{2ix}\right)^{k}\right)\\
&= \operatorname{Im}\left(e^{ix}\frac{e^{2nix}-1}{e^{2ix}-1}\right)\\
&= \operatorname{Im}\left(\frac{e^{2nix}-1}{e^{ix}-e^{-ix}}\right)\\
&= \operatorname{Im}\left(\frac{\cos(2nix)-1+i\sin(2\pi ix)}{2i\sin x}\right)\\
&= \frac{\sin(2nx)}{2\sin x}
\end{align}
Now it should be clear that it is bounded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Short method to evaluate $\lim_{x\to \frac\pi2} \frac{(1-\tan(\frac x2))(1-\sin(x))}{(1+\tan(\frac x2))(\pi-2x)^3}$? $$\lim_{x\to \frac\pi2} \frac{(1-\tan(\frac x2))(1-\sin(x))}{(1+\tan(\frac x2))(\pi-2x)^3}$$
I only know of L'hopital method but that is very long. Is there a shorter method to solve this?
| Another trick is to multiply both the numerator and denominator by
$(1+\tan(x/2))(1+\sin x)$ and use that
\begin{align}
1-\sin^2x&=\cos^2x,\\
1-\tan^2(x/2)&=\frac{\cos^2(x/2)-\sin^2(x/2)}{\cos^2(x/2)}=\frac{\cos x}{\cos^2(x/2)}.
\end{align}
Then you get after simplification
$$
\frac{1}{(1+\tan(x/2))^2(1+\sin x)\cos^2(x/2)}\frac{1}{2^3}\color{red}{\left(\frac{\cos x}{\pi/2-x}\right)^3}
$$
where the only uncertainty when $x\to\pi/2$ is in the red part. The limit of $\frac{\cos x}{\pi/2-x}$ can be calculated by L'Hopital.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059571",
"timestamp": "2023-03-29T00:00:00",
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Question on Hilbert Space $S$ is a non-empty subset of a Hilbert Space $H$, show that the set of all linear combinations of vectors in $S$ is dense in $H\Leftrightarrow S^{\bot}=\{0\}$
I have been able to show $\Rightarrow$ direction : Call the set of all linear combinations of vectors in $S$ as $S_1$ Say, $v\in S^{\bot}, \exists$ a sequence $v_n$ in $S_1$ such that $v_n\rightarrow v$ as $n\rightarrow\infty$, $v_n\bot v $
$\forall n\in\mathbf{N}$ $\therefore \lim_{n\rightarrow\infty}\|v_n-v\|^2=\lim_{n\rightarrow\infty} \|v_n\|^2+\|v\|^2=2\|v\|^2=0 \Rightarrow v=0$
I am struggling with the other implication, We have $S^{\bot\bot}=H$, need to show closure of $S_1$ is $S^{\bot\bot}=H$
| Notice that $\left(\overline{\operatorname{span}S}\right)^\perp = S^\perp$. The Riesz projection theorem implies $$\left(\overline{\operatorname{span}S}\right)\oplus \left(\overline{\operatorname{span}S}\right)^\perp = H$$
so $$\text{linear span of } S \text{ is dense in } H \iff\overline{\operatorname{span}S} = H \iff \left(\overline{\operatorname{span}S}\right)^\perp = \{0\} \iff S^\perp = \{0\}$$
| {
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If $A=A^2$ is then $A^T A = A$? I know that for a matrix $A$:
If $A^TA = A$ then $A=A^2$
but is it if and only if? I mean:
is this true that "If $A=A^2$ then $A^TA = A$"?
| Since $$\det (A)^2 = \det (A)\det (A) = \det (A^2)= \det (A) \implies \det (A)\in\{0,1\}$$
So if $\det (A)= 1$ then exsist $A^{-1}$ so $A = I$ and the answer is yes.
If $\det(A)=0$ then examples show that answer is negative.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_{x^2+y^2+z^2 \leq 1}\frac{dx\,dy\,dz}{x^2+y^2+(z-2)^2}$ I'm trying to calculate the integral $$\int_{x^2+y^2+z^2 \leq 1}\frac{dx\,dy\,dz}{x^2+y^2+(z-2)^2}.$$
I've tried in two methods:
Regular spherical coordinates, but this leads to really unfun integrals and logarithms with negative numbers inside them and other beasts I'd rather avoid.
The other method was to try shifted spherical coordinates, $x = r\sin(\theta)\cos(\phi), y=r\sin(\theta)\sin(\phi)$ but $z-2 = r\cos(\theta)$.
Now the integral is very easy but finding the limits of integration is tougher. We still have $0 < \theta < \pi$ and $0 < \phi < 2\pi$, I think, but the limits on $r$ are harder.
At the very limit of the domain of integration we have $x^2+y^2+z^2 = 1$, so $r^2+4r\cos(\theta)+4 = 1$, so we need to have $r^2+4r\cos(\theta) + 3 = 0$.
This happens when $$r = \frac{-4\cos(\theta)\pm\sqrt{16\cos^2(\theta)-12}}{2}.$$ I don't know how this helps us or if I'm going in the right direction.
The unfun way:
$\int_{x^2+y^2+z^2 \leq 1}\frac{dxdydz}{x^2+y^2+(z-2)^2} = \int_{0}^{1}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{r^2\sin(\theta)}{r^2-4r\cos(\theta)+4}d\phi d\theta dr = 2\pi \int_{0}^{1}\int_{0}^{\pi}\frac{r^2\sin(\theta)}{r^2-4r\cos(\theta)+4}d\theta dr $
Use $u-$substitution $u=\cos(\theta)$ to get:
$2\pi \int_{0}^{1}\int_{0}^{\pi}\frac{r^2\sin(\theta)}{r^2-4r\cos(\theta)+4}d\theta dr =-2\pi \int_{0}^{1}\int_{1}^{-1}\frac{r^2}{r^2-4ru+4}dudr = 2\pi\int_{0}^{1}\int_{-1}^{1}\frac{r^2}{r^2-4ru+r}dudr = 2\pi\int_{0}^{1}r^2[\frac{\ln(r^2-4ru+4)}{-4r}]_{-1}^{1}dr = -\frac{\pi}{2}\int_{0}^{1}r(\ln(r^2-4r+4) - \ln(r^2+4r+4))dr=-\pi\int_{0}^{1}r(\ln(|r-2|)-\ln(r+2))dr$
Is this the right way? seems very unpleasant...
| Consider the integral $$I(a, b) = \int_{x^2 + y^2 \le a} \frac{dx \, dy\, }{x^2 + y^2 + b}$$ By changing to polar coordinates, we compute $$I(a,b) = \int_{0}^{2\pi} \int_{0}^{\sqrt{a}} \frac{r \, dr\, d\theta}{r^2 + b} = \pi \ln(a/b + 1)$$ Your desired integral is $$\int_{-1}^{1} I(1-z^2, (z-2)^2) \, dz = \pi \int_{-1}^{1} \ln\left(\frac{1-z^2 + (z-2)^2}{(z-2)^{2}} \right)\, dz = \pi \int_{-1}^{1} \ln(5-4z) - 2\ln(|z-2|) \, dz$$ which one can compute by elementary techniques (e.g. integration by parts).
| {
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"timestamp": "2023-03-29T00:00:00",
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System of equations has no solution
If the system of linear equations,
\begin{cases}
x &+ ay &+ z &= 3\\
x &+ 2y &+ 2z &= 6\\
x &+ 5y &+ 3z &= b\\
\end{cases}
has no solution, then:
*
*$a=-1,b=9$
*$a=-1,b \ne 9$
*$a\ne-1,b = 9$
*$a=1,b \ne 9$
I really fail to understand why the answer cannot be option (1).
Here's my method:
For $a = -1$, the coefficient determinant determinant is 0. And if we create $\Delta_z$, the determinant in which coefficients of $z$ are replaced by $3,6,b$, (b= 9), that determinant is non zero.
My book says that if coefficient determinant is $0$ and at least one of $\Delta_y, \Delta_z, \Delta_x$ is non $0$ then the system has no solution.
But the answer given is $(2)$. What's wrong with my method? What's the best way to tackle such problems?
| I would do it this way, using row reduction. The criterion for a non-homogeneous linear system to have solutions is that the matrix of the linear system (l.h.s.) and the augmented matrix have the same rank.
From this criterion we deduce readily that if this system has no solution, the matrix of the l.h.s. cannot have rank $3$, hence its determinant, which is equal to $-(a+1)$ is zero, whence $a=-1$. Computing the determinant by row reduction shows this matrix has rank $2$.
Now, if $a=-1$, the augmented matrix reduces as follows:
$$\begin{bmatrix}1&-1&1&3\\1&2&2&6\\1&5&3&b\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1&-1&1&3\\0&3&1&3\\0&3&1&b-6\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1&-1&1&3\\0&3&1&3\\0&0&0&b-9\end{bmatrix} $$
Thus, for $a=-1$, the augmented matrix has rank $3$ if and only if $b-9\ne 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Defining the Cosine Function from First Principles, intuitively Throughout most of my mathematics education, the cosine function has been defined formally either using its series expansion, as $\mathfrak{Re}(\exp i\theta)$, or as the unique solution to $y+y''=0$ with $y(0)=1$ and $y'(0) = 0$.
Although these definitions make the mathematics more convenient, I've always been interested to see what it would be like to try to formalise it directly with the geometric intuition from the unit circle, assuming only the notion of distance (i.e. that we have a complete metric space with metric $d\colon\mathbb R^2 \times \mathbb R^2 \to \mathbb R$).
Since, by geometric intuition, we have that the cosine function is the $x$-coordinate obtained by walking along the unit circle, my idea was to take a line segment from the point $(1,0)$ to another point on the circle such that the distance is $\theta$, and split it into two line segments whose sum of lengths is also $\theta$. We then split into 3 line segments, and so on, in a limiting manner,
in such a way that the sum of lengths is always $\theta$, as illustrated below with $\theta = 2\pi/5$.
[Desmos link]
When this procedure converges, the $x$- and $y$-coordinates of the limiting point should be the cosine and sine of $\theta$ respectively. The problem is I can't find an easy way to express the $n$th coordinate of this procedure (assuming we are splitting the line into $n-1$ segments) so that I can take the limit.
I was going to try and do things the other way round, that is, define $\cos^{-1}\colon [-1,1] \to [0,\pi]$ by approaching the circle from below and finding the length. I did manage to do this, and got that
\begin{align*}
\cos^{-1}(x) &= \sum_{k=1}^\infty d\left[\left(\frac{(k-1)x}{n}, \sqrt{1-\left(\frac{(k-1)x}{n}\right)^2}\right), \left(\frac{kx}{n}, \sqrt{1-\left(\frac{kx}{n}\right)^2}\right)\right]\\[4pt]
&= \sum_{k=1}^\infty \sqrt{2
\left(1-\sqrt{1-\frac{(k-1)^2
x^2}{n^2}}
\sqrt{1-\frac{k^2
x^2}{n^2}}\right)-\frac{2
k(k-1) x^2}{n^2}},
\end{align*}
and I suppose one can extrapolate from here to obtain the $\cos$ and $\sin$ functions and extend them appropriately.
But it feels like the other way, that is, finding that limiting point, should be possible. I appreciate any assistance with this.
| Let $\alpha_n$ be the central angle of the first trangle in the $n$-th iteration.
It is an isosceles triangle with base length $\frac{\theta}n$ and leg length $1$. Hence $\alpha_n$ is given by the equation
$$\sin\frac{\alpha_n}2 = \frac{\theta}{2n}$$
or $\alpha_n = 2\arcsin\left(\frac{\theta}{2n}\right)$.
The coordinates of the last ($n$-th) point are then
$$(\cos (n\alpha_n), \sin (n\alpha_n)) = \left(\cos\left(2n\arcsin\left(\frac{\theta}{2n}\right)\right), \sin\left(2n\arcsin\left(\frac{\theta}{2n}\right)\right)\right) \xrightarrow{n\to\infty} (\cos\theta, \sin\theta)$$
because of the limit $\lim_{t\to 0} \frac{\arcsin t}t = 1$ and the continuity of $\sin$ and $\cos$.
This shows that your last constructed point indeed converges to $(\cos\theta, \sin\theta)$. It doesn't give an algebraic expression for the coordinates, though.
| {
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Rationalizing denominator of $\frac{18}{\sqrt{162}}$. Cannot match textbook solution I am given this expression and asked to simplify by rationalizing the denominator:
$$\frac{18}{\sqrt{162}}$$
The solution is provided:
$\sqrt{2}$
I arrived at:
$$\frac{\sqrt{162}}{9}$$
Here is my thought process to arrive at this incorrect answer:
$\frac{18}{\sqrt{162}}$
= $\frac{18}{\sqrt{162}}$ * $\frac{\sqrt{162}}{\sqrt{162}}$
= $\frac{18\sqrt{162}}{162}$
= $\frac{\sqrt{162}}{9}$
How can I arrive at $\sqrt{2}$ ?
| $\frac{\sqrt{162}}{9} = \frac{\sqrt{2 \cdot 9^2}}{9} = \frac{9\sqrt{2}}{9} = \sqrt{2}$.
| {
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Can I conclude that $\lim_{x\to0^+}\frac{x^2}{e^{-\frac{1}{x^2}}\cos(\frac{1}{x^2})^2}$ is infinite or it doesn't exists? $$\lim_{x\to0^+}\frac{x^2}{e^{-\frac{1}{x^2}}\cos(\frac{1}{x^2})^2}$$
My intuition is that the denominator goes to 0 faster and everything is non-negative, so the limit is positive infinity.
I cant think of elementary proof, l'Hopital doesn't help here, and I'm not sure if and how to use taylor.
WolframAlpha says it's complex infinity, but I can't understand why it doesn't just real positive infinity.
I've tried to use wolfram language to use the assumption that x is real, but couldn't get any result.
| The limit$$\lim_{x\to0^+}\frac{x^2}{\exp\left(-\frac1{x^2}\right)}$$is indeed $+\infty$. Since $\dfrac1{\cos^2\left(\frac1{x^2}\right)}\geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+\infty$.
| {
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Find the orthogonal bases of the space $ V $ and $ V^{\perp}$ In space $\mathbb R^3 $ f Find the orthogonal bases of the space $ V $ and $ V^{\perp}$ where $$V = \left\{ \vec{x} \in \mathbb R^3 : x_1 - 3x_2 + x_3 = 0 \right\} $$
On the begining, I know that may I ask you for basics of topic, but I truly have some doubts...
Ok, I found basis of $V$: $ [3,1,0]^T,[-1,0,1]^T$ and I have used Gram–Schmidt process to find orthogonal basis of $V$:
$[3,1,0]^T,[-1/10,3/10,1]^T$
Ok, now I should find basis of $ V^{\perp} $ and on the same algorithm find orthogonal basis of $V^{\perp}$ - but there is my question - if in basis of $V^{\perp}$ I have all vectors which are perpendicular to vectors from basis $V$, the basis of $ V^{\perp} $ will be just orthogonal basis of $V$? And orthogonal basis of $V^{\perp}$ will be just basis of $V$? Have I right or there is something other to do?
| No, it's wrong. If you have an orthogonal basis $\mathcal{B} = \{v_1,\cdots, v_n\}$ of $V$, then for each $v_i$, it is perpendicular to other $v_j$'s but not to itself. So $v_i\notin V^\perp$ for all $i$. By the way, you can find an orthonormal basis of $V^\perp$ easily from the definition of $V$ as
$$
V = \{(1,-3,1)\}^\perp.
$$ (It's just $\{\frac{1}{\sqrt{11}}(1,-3,1)'\}$.)
| {
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Limit of $\frac{1}{7}e^{-2x^2}(1-4x^2)$ as $x\to\infty$ I calculated the derivative of $\frac{x}{7}*e^{-2x^2}$ and got $\frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)
I don't know how to find the limit of this function:
$$\frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$\lim_{x \to \infty} \frac{1}{7}e^{-2x^2}+\lim_{x \to \infty} \frac{1}{7}xe^{-2x^2}*(-4x) $$
| Hint: Rewrite the expression as
$$\frac{1}{7}e^{-2x^2}\left(1-4x^2\right) = \frac{1-4x^2}{7e^{2x^2}}$$
Now, notice the growth of the numerator and denominator. Which grows more quickly?
| {
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Any good way to calculate $\frac {\alpha ^ n - 1 } {\alpha - 1} \pmod{c}$ I tried by multiplying modular inverse of denominator to the numerator and then taking modulo $c$, but there are problems when the inverse does not exist.
So is there a good way to solve this problem.
Constraints
$$ 1 \le \alpha \le 1e9 $$
$c$ is a prime
$$ 1 \le n \le 1e9 $$
| Set $S_0:=1$ and then recursively $S_k:=\alpha S_{k-1}+1 \pmod c$ for all $k=1,\dotsc,n-1$. The last value $S_{n-1}$ is what you seek.
| {
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calculus and series If we suppose $\lim_{n\to+\infty} a_n =0,$ I have to examine if $\sum _{ n=1 }^{ \infty } n^{\frac {1}{1+a_n}}$ converges or diverges.
I used the monotone convergence theorem to show that the series diverges. If we suppose: $a_n=\dfrac {1}{n},$ then $$\lim_{n\to+\infty} a_n= \lim_{n\to+\infty} \frac {1}{n}=0. $$ Let the sequence of functions $f$ on $\mathbb{N}$ be defined by $$f_n (k)= k^{\frac{1}{1+\frac {1}{n}}}.$$ We note that these functions are positive, and that for each $k$, as a sequence in $n, f_n (k)$ is increasing. The sum over $k$ is an integral and it's the integral over $\mathbb{N}$ with respect to counting measure. In particular, by the Monotone Convergence Theorem (from the Lebesgue theory) allows to interchange limit and integral:
$$\lim_{n\to+\infty} \sum _{ k=1 }^{ \infty } k^{\frac {1}{1+\frac {1}{n}}} = \sum _{ k=1 }^{ \infty } \lim_{n\to+\infty} k^{\frac {1}{1+\frac {1}{n}}}= \sum _{ k=1 }^{ \infty } \frac{1}{k}= +\infty$$
and then $$ \sum _{ k=1 }^{ \infty } k^{\frac {1}{1+\frac {1}{n}}} $$ diverges. By the same argument, $$\sum _{ k=1 }^{ \infty } k^{\frac {1}{1+a_n}}$$ diverges.
| Hint: It is absolutely diverging because $n^x$ is equal or greater than $1$ for $n \geq 1$ and $x \geq 0$.
| {
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Proof explanation of $``\exists x\in\mathbb{R}$ with $x^2=2"$ Can someone please help me break down the proof below from $(*)$ onwards. I'm lost at what is going on and where the proceeding steps are coming from. Is this a proof by contradiction? Why are we assuming $M^2\lt 2 , M^2\gt 2$, and choosing $\delta$ to be the minimum of $\frac{2-M^2}{5}$ and $1$?
Theorem 2.13
(Square root 2 exists). There exists $x\in\mathbb{R}$ with $x^2=2.$
Proof.
Define $$A=\{y\in\mathbb{R}|y^2 \leq{2}\}.$$
As $0^2\leq{2}$, we have $0\in{A}$, so $A$ is non-empty. Now suppose $y\in\mathbb{R}$ has $y\geq{2}$, then $y^2\geq{4}$, so $y\notin{A}$. Thus, $2$ is an upper bound for A and hence A is bounded above. Therefore, by the completeness axiom, $M=\sup(A)$ exists. Note that $M\geq{1}$ as $1\in{A}$ and $M\leq{2}$ is an upper bound for A.
$(*)$
Suppose $M^2\lt{2}$. Choose $\delta\gt{0}$ with $\delta\lt $ min$(\frac{2-M^2}{5},1)$, and then define $ y=M+\delta$. As $\delta\lt{1}$, we have $\delta^2 \lt{\delta}$ and so
$$y^2=(M+\delta)^2=M^2+2M\delta+\delta^2\leq M^2+4\delta+\delta=M^2+5\delta\lt 2.$$
Thus $y\in A$, but this is a contradiction as $y\gt M$. Therefore $M^2\geq 2$.
Suppose $M^2\gt 2$. Choose $\delta \gt 0$ with $\delta\lt $min$(M,\frac{M^2-2}{2M})$ so that $M^2-2M\delta \gt 2$ and $M-\delta\gt 0$. Then,
$$(M-\delta)^2=M^2-2M\delta+\delta^2\geq M^2-2M\delta\gt 2.$$
Now if $y\geq (M-\delta)$, then $y^2\geq (M-\delta)^2 \gt 2$ so $y\notin A$. Thus $M-\delta$ is an upper bound for $A$, contradicting the fact that $M$ is the supremum of $A$
Since $M^2 \lt 2$ and $M^2\gt 2$ both lead to contradictions, we conclude that $M^2=2$, as required.
| Do you know the trichotomy law of inequality? It says that for any two numbers $a,b$, either $a < b$, or $a > b$, or $a=b$.
So now if you can prove that $a<b$ leads to a contradiction, and that $a > b$ leads to a contradiction, the only possibility left must be true, namely $a=b$. Apply this with $a=M^2$ and $b=2$.
| {
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Show that it doesn't exist any homomorphism $\phi: \mathbb{C} \to \mathbb{R}^{\mathbb{R}}$
*
*Considering the ring of all of the real functions $ \mathbb{R}^{\mathbb{R}}$. Show that it doesn't exist any homomorphism $\phi: \mathbb{C} \to \mathbb{R}^{\mathbb{R}}$
*Let $R$ be an integral domain and consider $f:R\to R\,\,\,\,$ s.t. $\,\,\,\, f(a)=a^2,\,\,\,\,\forall a \in R$. Show that $f$ is injective $\iff$ $f$ is a homomorphism.
About 1. any hint on how to show that?
About the point 2.: I have been thinking about that $\forall a \in R, \,\,\, f(a) = a^2 = (-a)^2 = f(-a)$ and, assuming $f$ injective, it means $a=-a \Rightarrow a + a = 2a =0_R\,\,\,\,$ Then: $\,\,\,\,\,a^2+b^2 = a^2+ (2a)b +b^2=(a+b)^2=f((a+b))=f(a)+f(b)=a^2+b^2\,\,\,\,\,$
The fact that $R$ is an I.D. gives the product because we can say that $(ab)^2=a^2b^2$, so $f$ is homomorphism. To show that homomorphism implies injectivity do I need only to retrace this demonstration?
| For your other question: if $-1 \neq 1$, then $f $ can't be injective since $f(-1) = f(1) = 1$. And if $-1 = 1$, then $2=0$ and so $f$ respects addition as per your calculation.
| {
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Find all 2 by 2 complex matrix with the following condition
Find all $A \in \mathrm{Mat}_{2 \times 2} ( \mathbb{C})$ such that
$A^2 = -I$ and prove that there is no 2 by 2 real diagonal matrices
$A$ with $A^2 = -I$. Deduce that for every even $n$ there are
infinitely many $n \times n$ real matrices with $A^2 = -I$.
Try:
We have
$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \implies A^2 = \begin{bmatrix}a^2+bc & b(a+d) \\ c(a+d) & cb+d^2 \end{bmatrix} $$
So $A^2 = - I$ iff $-1 = a^2+bc = cb + d^2$ and $0=b(a+d)=c(a+d)$
If $b=0$, then we have $a^2=-1 $ , $d^2=-1$.
if $a^2=-1$, then $a= \pm i$ and similarly $b= \pm i$. As long as $a+d \neq 0$, we see that $c=0$ otherwise $c \in \mathbb{C}$ is arbitrary and so we have
$$ \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}, \begin{bmatrix} -i & 0 \\ 0 & -i \end{bmatrix},\begin{bmatrix} -i & 0 \\ c & i \end{bmatrix},\begin{bmatrix} i & 0 \\ c & -i \end{bmatrix} $$
now if $b \neq 0$, then $a+d=0$ and so $a=-d$ and thus $c = -(1+a^2)/b$
Therefore, we have
$$ \begin{bmatrix} a & b \\ -(1+a^2)/b & -a \end{bmatrix} $$
are all posible matrices. Now, if the $a,b,c,d$ were real, we see that $b \neq 0$ and so we get matrices like the one above only.
Now, is this correct so far? I'm stuck for the $n\times n$ case. Any help?
| Looking very good so far. However, for the $a=\pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.
For even $n$, take the real $2\times2$ matrix you've found, make $\frac n2$ copies of it all along the diagonal of an $n\times n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $n\times n$ matrices.
Bonus question: What happens for odd $n$ with real matrices?
| {
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Uncountable sum of vectors in a Hilbert Space I am currently reading Hilbert Spaces and confused about a thing. Say, $C=\{e_\alpha : \alpha\in\mathcal{A}\}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $\sum_{\alpha\in\mathcal{A}}e_\alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?
| The sum $\sum_{\alpha\in A} v_\alpha$ does make sense. It is defined to converge to $L \in H$ if
$$\forall \varepsilon > 0 \,\exists F_0 \subseteq A \text{ finite such that }\forall F \subseteq A \text{ finite}, F \supseteq F_0 \text{ we have} \left\|\sum_{\alpha\in F}v_\alpha - L\right\| < \varepsilon$$
However, the sum $\sum_{\alpha\in A} e_\alpha$ of an orthonormal set only converges if $A$ is finite.
Indeed, assume that $\sum_{\alpha\in A} e_\alpha = L$. For $\varepsilon = \frac12$ there exists $F_0 \subseteq A$ finite such that for all $F \subseteq A$ finite, $F \supseteq F_0$ implies $\left\|\sum_{\alpha\in F} e_\alpha -L\right\| < \frac12$.
For any $F_0 \subseteq F \subseteq A$ finite we have
$$\sqrt{|F \setminus F_0|}= \left\|\sum_{\alpha\in F\setminus F_0}e_\alpha\right\| = \left\|\sum_{\alpha\in F}e_\alpha - \sum_{\alpha\in F_0}e_\alpha\right\| \le
\left\|\sum_{\alpha\in F}e_\alpha - L\right\| +\left\|L- \sum_{\alpha\in F_0}e_\alpha\right\| < 1$$
so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.
| {
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Proving the Fibonacci identity $(−1)^{m−k}(F_{m+k+1}F_{m−k−1}−F_{m+k}F_{m−k}) =F_{k}^2+F_{k+1}^2$ Prove that for two natural numbers $m$ and $k$, where $m>k$ the following identity holds:
$$(−1)^{m−k}(F_{m+k+1}F_{m−k−1}−F_{m+k}F_{m−k}) =F_{k}^2+F_{k+1}^2$$
Here the exercise comes with a hint:
The constant is $F^2 _m$, try to substitute $k=0$ in.
Okay fair enough, let us try this, this will give us:
$$ (−1)^{m}(F_{m+1}F_{m−1}−F_{m}F_{m}) =F_{0}^2+F_{1}^2$$
$$ (−1)^{m}(F_{m+1}F_{m−1}−F_{m}^2) =F_{0}^2+F_{1}^2=0^2+1^2=1$$
We can now use a identity by Cassini, namely that for $m>0=k$ we have:
$$ (−1)^{m}(F_{m+1}F_{m−1}−F_{m}^2) =(-1)^m(-1)^m=(-1)^{2m}=1 \checkmark$$
Okay great, so our identity works whenever $k=0$, but I do not see how this helps with a generalisation.
Edit/idea: Did I just write down the base case for an approach via induction on $k$ perhaps?
| Here is a proof based upon Binets formula
\begin{align*}
F_k=\frac{\varphi^k-\psi^k}{\varphi-\psi}\qquad k\geq 0\tag{1}
\end{align*}
where $\varphi=\frac{1}{2}\left(1+\sqrt{5}\right), \psi=\frac{1}{2}\left(1-\sqrt{5}\right)=-1/\varphi$.
We start with the right-hand side of OPs formula and obtain
\begin{align*}
F_k^2=\left(\frac{\varphi^k-\psi^k}{\varphi-\psi}\right)^2&=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}-2\left(\varphi\psi\right)^k+\psi^{2k}\right)\\
&=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}-2(-1)^k+\psi^{2k}\right)\tag{2}\\
\end{align*}
Putting $k\to k+1$ in (2) we get
\begin{align*}
\color{blue}{F_k^2+F_{k+1}^2}&=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}-2(-1)^k+\psi^{2k}\right)\\
&\qquad+\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k+2}-2(-1)^k+\psi^{2k+2}\right)\\
&\,\,\color{blue}{=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}\left(1+\varphi^2\right)+\psi^{2k}\left(1+\psi^2\right)\right)}\tag{3}
\end{align*}
And now the left-hand side.
We obtain
\begin{align*}
F_{m+k}F_{m-k}&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{m+k}-\psi^{m+k}\right)\left(\varphi^{m-k}-\psi^{m-k}\right)\\
&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}-\varphi^{m+k}\psi^{m+k}-\varphi^{m-k}\psi^{m+k}+\psi^{2m}\right)\\
&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}-\left(\varphi\psi\right)^{m-k}\left(\varphi^{2k}+\psi^{2k}\right)+\psi^{2m}\right)\\
&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}-(-1)^{m-k}\left(\varphi^{2k}+\psi^{2k}\right)\right)\tag{4}
\end{align*}
Replacing $k$ with $k+1$ in (4) we get
\begin{align*}
F_{m+k+1}F_{m-k-1}&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}+(-1)^{m-k}\left(\varphi^{2k+2}+\psi^{2k+2}\right)\right)\tag{5}
\end{align*}
The left-hand side of OPs formula can now be rewritten using (4) and (5) as
\begin{align*}
\color{blue}{(-1)^{m-k}}&\color{blue}{\left(F_{m+k+1}F_{m-k-1}-F_{m+k}F_{m-k}\right)}\\
&=\frac{(-1)^{m-k}}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}+(-1)^{m-k}\left(\varphi^{2k+2}+\psi^{2k+2}\right)\right)\\
&\qquad-\frac{(-1)^{m-k}}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}-(-1)^{m-k}\left(\varphi^{2k}+\psi^{2k}\right)\right)\\
&\,\,\color{blue}{=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}\left(1+\varphi^2\right)+\psi^{2k}\left(1+\psi^2\right)\right)}\tag{6}
\end{align*}
A comparison of (3) and (6) shows OPs identity is valid.
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Limit of $x_n=\sum_{k=np+1}^{nq}\frac{1}{k}$ using Riemann sum I am trying to find the limit of the following sequence using Riemann sum:
$$x_n=\sum_{k=np+1}^{nq}\frac{1}{k}\qquad p,q\in\mathbb{N}\quad p<q$$
I have tried to develope the expression:
$$\frac{1}{np+1}+\frac{1}{np+2}+...+\frac{1}{nq}=\frac{1}{n}(\frac{1}{p+\frac{1}{n}}+\frac{1}{p+\frac{2}{n}}+...+\frac{1}{p+\frac{n(q-p)}{n}})=\frac{1}{n}\sum_{k=1}^{n(q-p)}\frac{1}{p+\frac{k}{n}}$$
But I need an expression like $\frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})$, from $k=1$ to $n$, not to $n(q-p)$, so I can calculate the limit as $\int_0^1f(x)dx.$
Could you give me some hints? Thanks!
| We just have to calculate $\int_0^{q-p}$ instead of $\int_0^1$. Then, with
$$f(x)=\frac{1}{p+x}$$
$$\int_0^{q-p}\frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=ln\frac{q}{p}$$
And we are done.
| {
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If $|G|=p^2$, $p$ prime and $G$ is not cyslic $G\cong \Bbb Z/p\Bbb Z\times \Bbb Z/p\Bbb Z $ Here is an incomplete proof:
If $H\le G$ is a subgroup, there are only three possibilities for its cardinal: $1,p$ or $p^2$ by Lagrange.
If $|H|=p^2$ then there exists $g\in H\subset G$ of order $p^2$ and so $G$ is cyclic.
If $|H|=1$ and there is no other proper subgroup then $G$ is cyclic.
So there must be a subgroup of order $p$, let $|H|=p$. Let $x$ be in $G$. We will show that $xHx^{-1}=H$ and so $H\triangleleft G.$ By contradiction let's suppose $xHx^{-1}\ne H$.
$xHx^{-1}$ is a subgroup of $G$ because $$(xhx^{-1})^{-1}=xhx^{-1}\in xHx^{-1} \text{ and }~ xhx^{-1}xh'x^{-1}=xhh'x^{-1}\in xHx^{-1}~\forall h,h'\in xHx^{-1}$$
There is a formula for the cardinality of a product of groups that says $$\begin{align}|xHx^{-1}|\cdot|H|&= |xHx^{-1}H|\cdot|xHx^{-1}\cap H|\\ |(xHx^{-1})H| &= \frac{|xHx^{-1}|\cdot|H|}{|xHx^{-1}\cap H|}=\frac{p\cdot p}{1}\end{align}$$
And here I'm not sure how to show that $xHx^{-1}$ has cardinality $p$, I guess for the same reason as $H$.
And why is the intersection trivial?
If the equality above is true we get that $xHx^{-1}H=G$ so for any $x\notin H$ there exist some $h_1,h_2\in H$ such that $x=xh_1x^{-1}h_2$
$$\begin{align}e &=h_1x^{-1}h_2\\ h_1^{-1}&= x^{-1}h_2\\ h_1^{-1}h_2^{-1}&= x^{-1}\implies x\in H \text{ (impossible)}\end{align}$$
Once we arrive at a contradiction we have that $xHx^{-1}=H$ so $H$ is normal in $G$.
Then I have to show that there exists $K\le G$ such that $K\cap H=\{e\}$ and $K\triangleleft G$ and from this it will be easy to show that $G\cong K\times H$
| $H$ has cardinality $p$. The map $H\to xHx^{-1}$ defined by $h\to xhx^{-1}$ is one to one and onto. I'll leave you to check that, it is very easy. Hence $xHx^{-1}$ must have the same cardinality. As for the intersection there are only two options: it is either of size 1 or $p$. If it has size $p$ then it means that $H$ and $xHx^{-1}$ is the same set which contradicts your assumption.
| {
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"timestamp": "2023-03-29T00:00:00",
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Index of summation - integer? Can the index of summation of a sum be anything other than an integer? What is the reasoning behind the answer?
| The "index" of summation, the way I'm interpreting your question, corresponds to an enumeration of the elements of the set over which you are summing. For example, if you want to add the numbers $x_{1},...,x_{n}$, then you would write
$$\sum_{k=1}^{n}x_{k}$$
You can also think of this as
$$\sum_{x\in S}x$$
where $S=\{x_{1},...,x_{n}\}$.
However, there are sets which are not countable - this means the elements of the set are not in one-to-one correspondence with the set of integers. For example, you can take the sum of every real number in $[0,1]$, which you could write as
$$\sum_{x\in [0,1]}x$$
The problem is that because this set is uncountable, so there's no natural ordering for which to add these numbers. And in this case the sum is infinite. However if you know that a sum
$$\sum_{x\in S}x<\infty$$
then, even if $S$ is uncountable, this can be rewritten as a sum over a countable set.
| {
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In $\Delta ABC$ if $(\sqrt{3}-1)a=2b$, $A=3B$, then find $C$
In $\Delta ABC$ if $(\sqrt{3}-1)a=2b$, $A=3B$, then find $C$
My Attempt
$$
b=\frac{\sqrt{3}-1}{2}a\quad\& \quad \frac{A-B}{2}=B\quad\&\quad\frac{A+B}{2}=2B\\\frac{a-b}{a+b}=\frac{\tan\frac{A-B}{2}}{\tan\frac{A+B}{2}}\implies \frac{3-\sqrt{3}}{\sqrt{3}+1}=\frac{\tan B}{\tan 2B}=\frac{1-\tan^2B}{2}
$$
| $$2\sin B=(\sqrt3-1)\sin3B=(...)(\sin B)(3-4\sin^2B)$$
As $\sin B>0,$ $$\sin^2B=\dfrac{2-\sqrt3}4$$
$$\cos2B=1-\dfrac{2-\sqrt3}2=\cos30^\circ$$
$0<2B<360^\circ,2B=360^\circ n\pm30^\circ$ for some integer $n$
$\implies2B=30^\circ$
| {
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If $f \circ f = 0 $, show that transformations $f + id_x$ and $f - id_x$ are isomorphisms of $X$ I have a problem with this task:
The linear transformation $f \in L(X,X)$ has property $f \circ f = 0 $
Show that transformations $f + id_x$ and $f - id_x$ are isomorphisms of $X$ space with itself
If I need to be honest I have no idea how to prove this. I was tryinging something like that:
If $f + id_x$ is isomorphism it must be both injective and surjective.
Ok, but $id_x$ is injective and surjective.
I thought to show that f is surjective but unfortunately sum of two surjectives can give me something what is not surjective...
Maybe the key is in $f \circ f = 0 $?
Thanks for your time!
| As all the other answers have amended my comment by actually going through the computation, I just want to add two levels of generalisations of the result which I see:
i) We only need the abelian group structure on $X$. In other words, with the same proof we get the more general result:
If $A$ is an abelian group and $f \in End(A)$ satisfies $f\circ f =0$, then $\pm id_A \pm f$ are automorphisms of $A$.
ii) Now the $End(A)$ in part i is a unital (but not necessarily commutative) ring $R$ with addition as addition and composition as multiplication, $id_A =1_R$. From this perspective, the result is just a special case of
If $u \in R^\times$, $n \in R$ is nilpotent, and $u$ and $n$ commute with each other, then $u+n \in R^\times$.
which is proved by a geometric series argument as suggested in Behnam Esmayli's comment. In our special case $n=\pm f$ with $n^2=0$, of course the inverse of $u+n$ is just $u^{-1}(1-n)$. See Units and Nilpotents and its many duplicates for a general discussion.
| {
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Is right this chromatic polynomial for this Bridge Graph? I have the following graph:
Bridge Graph with N = 8
And I need to find its chromatic polynomial.
Based in my notes, I have reached the following result:
$$Pg(x) = \frac{((x-1)^4 + (x-1))^2}{x(x-1)} $$.
It is because I have applied this formula:
$$Pg(x) = \frac{Pg1(x) Pg2(x)}{Pkr(x)} $$.
Taking the Bridge Graph and the previous formula, I have replaced with:
$$ Pg1(x) = Pg2(x) = Pc4(x)= (x-1)^4+(x-1) $$
Because both sub-graphs are cycles.
On the other hand, the Pkr(x) in this case is equals to Pk2(x). So, the chromatic polynomial for a complete graph with n = 2, could be expressed as :
$$x(x-1)$$
So, my main doubt is if this reasoning is right. Because I'm not sure about my resolution and I think that Ia have missing something in the final response.
Thanks a lot!
| Using standard techniques:
$$x^8-9 x^7+36 x^6-82 x^5+114 x^4-96 x^3+45 x^2-9 x$$
| {
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Theorem 5.8 Baby Rudin. Some questions This is with reference to the page number 107 of Baby Rudin.
Theorem 5.8 Let $f$ be defined on $[a,b]$.If a point $x\in(a,b)$ is a
local maximum of function $f$, and if $f'(x)$ exist, then $f'(x)=0$.
I have some questions. Thanks in advance for reading and helping out.
*
*What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?
*The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function function with local maximum/minimum at some point say $p$ but derivative undefined at the point.
Edits:
We can take $f(x)=x^2$ on $[-1,1]$. $f'(0)=0$ but $0$ is not local maximum. I think this works for (1).
| Answer to question 1
The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.
Answer to question 2
Take $g(x)=\vert x \vert$, again look at zero.
| {
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Does $\int_0^1 f(x)=\int_0^1 xf(x)$ imply $\int_0^x f(t)$ has a root My question is whether or not the following is true:
If $f:[0,1]\to \mathbb{R}$ is a continous function such that
$$\int_0^1 f(x)dx=\int_0^1 xf(x)dx$$
then there exist $c\in(0,1)$ such that
$$\int_0^c f(x)dx=0$$
It is quite clear that in the interval $(0,1)$ there must be some points $a$ such that $f(1)f(a)<0$ but I can't say from this if the statement above is true or not.
| Hint: Note that
$$
\int_0^1 F(x)dx =\int_0^1\left(\int_0^x f(t)dt\right)dx=\int_0^1\left(\int_t^1dx\right)f(t)dt= \int_0^1 (1-t)f(t)dt
$$ where $F(x) = \int_0^x f(t)dt$.
| {
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$f \in C_{00}(\mathbb{R^p},\mathbb{C})$. $ \mapsto f_t \in L_\infty(\mathbb{R}^p, \mathcal B_p, \lambda_p, \mathbb{C})$ uniformly continuous? Continuing from here
Let $f_t(x):=f(x+t)$
Consider $f \mapsto f_t$ which is a linear, isometric bijection from $L_\infty(\mathbb{R}^p, \mathcal B_p, \lambda_p, \mathbb{C})\to L_\infty(\mathbb{R}^p, \mathcal B_p, \lambda_p, \mathbb{C})$ for every $t \in \mathbb{R}^p$
How can I show that for $f \in C_{00}(\mathbb{R^p},\mathbb{C})$ (continuous and compact support) the mapping $\mathbb{R}^p\ni t \mapsto f_t \in L_\infty(\mathbb{R}^p, \mathcal B_p, \lambda_p, \mathbb{C})$ is uniformly continuous?
| This follows from the fact that a continuous function with compact support is uniformly continuous. For a fixed $\varepsilon$, there exists $\delta$ such that if $s,t\in\mathbb R^p$ satisfy $\lVert t-s\rVert\lt\delta$, then $\left\lvert f(t)-f(s)\right\rvert\lt\varepsilon$.
For any $x\in\mathbb R^p$ and $s,t$ satisfying $\lVert t-s\rVert\lt\delta$, the following inequality holds
$$
\left\lvert f(x+t)-f(x+s)\right\rvert\lt\varepsilon
$$
hence
$$\left\lVert f_t-f_s\right\rVert_\infty ... $$
| {
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Liouville's theorem for harmonic functions I was reading the proof of Liouville's theorem for harmonic functions (in $\mathbb{R}^n$) in Wikipedia, but I could not understand where do they use in that proof the assumption that $f$ is bounded.
The proof -
Taken from - https://en.m.wikipedia.org/wiki/Harmonic_function
| The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $\infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.
| {
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Simplifying $s=w_1-c$, where $c=\frac{1}{2}(w_1-t_1+\frac{w_2-t_2}{R})$ I have to solve this easy equation, but can't find the same answer as the correction of the exercise.
We have two equations that I want to simplify (plugging $c$ into $s$)
$$c=\dfrac{1}{2}\left(w_1-t_1+\dfrac{w_2-t_2}{R}\right)$$
$$s=w_1-c$$
I find:
$$s=w_1-\dfrac{1}{2}\left(w_1-t_1+\dfrac{w_2-t_2}{R}\right)=\dfrac{1}{2}\left(w_1+t_1-\dfrac{w_2+t_2}{R}\right)$$
The correction actually says:
$$s=w_1-\dfrac{1}{2}\left(w_1-t_1+\dfrac{w_2-t_2}{R}\right)=\dfrac{1}{2}\left(w_1-t_1-\dfrac{w_2-t_2}{R}\right)$$
I must be wrong but I don't know why. Could somebody explain this to me?
| You first expand a bit to simplify for the coefficient of $w_1$:
$$w_1-\frac{1}{2}\left(w_1-t_1+\frac{w_2-t_2}{R}\right) = w_1-\frac{1}{2}w_1-\frac{1}{2}\left(-t_1+\frac{w_2-t_2}{R}\right)$$
$$= \frac{1}{2}w_1-\frac{1}{2}\left(-t_1+\frac{w_2-t_2}{R}\right)$$
Factoring $\frac{1}{2}w_1$ by $-\frac{1}{2}$, the expression simplifies to:
$$-\frac{1}{2}\left(-w_1-t_1+\frac{w_2-t_2}{R}\right)$$
And factoring by $-1$ yields:
$$\frac{1}{2}\left(\color{blue}{+}w_1\color{blue}{+}t_1\color{blue}{-}\frac{w_2-t_2}{R}\right)$$
You apparently made a mistake in the numerator of the final fraction. It should be $w_2-t_2$ rather than $w_2+t_2$, but I think it may have been a typo or an accident on your part. The rest is correct.
| {
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Integral Operator in $L^2$ I was trying to do this exercise and I'm wondering if I figured it out well:
I have $\mathcal{H} := L^2(0,1)$ and $T$ the operator with integral kernel $K(x,y) = \min\{x,y\}$, $x,y \in [0,1]$. I have to show that $T$ is compact and self-adjoint.
To show that is compact I was thinking to say that because $\min\{x,y\} \in [0,1]$ then
\begin{equation}
\dim(\operatorname{Im}T) = 1
\end{equation}
(The self adjointness I think is trivial..)So T belongs to finite rank operators and so it is compact. (Is this correct?) Then it asks me to find eigenvalues and eigenvectors of $T$ and here I really don't know how to proceed...
| We have
$$
\int_{(0,1)^2} |k(x,y)|^2\ \mathsf d(x\times y) = \int_0^1\int_0^1 (x\wedge y)^2\ \mathsf dx\ \mathsf dy
\leqslant \int_0^1\int_0^1\ \mathsf dx\ \mathsf dy = 1 <\infty,
$$
so T is a Hilbert-Schmidt operator and hence is compact.
| {
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Sum of set bits in every element for a natural numbers I was thinking of a mathematical puzzle with binary representation of numbers, but could not find a convincing answer myself.
Here is the puzzle: Say for some number N, I want to find the sum of the set bits of every number from 1 to N.
For example, for 5
The answer would be: 7 by the following procedure
1 - 1 set bit
2 - 1 set bit
3 - 2 set bits
4 - 1 set bit
5 - 2 set bits
So answer is 1 + 1 + 2 + 1 + 2 = 7
I found that it's easy to just go one by one and add, like I did. I also found that for a number having x bits, they form the pascal triangle with set ones, if number of occurances are counted with same number of set bits, irrespective of value. For example,
when x = 1, we have {1} - 1 set bit occurs once, hence 1.
when x = 2, we have {10, 11} - 1 set bit occurs once, 2 set bits occurs once, hence 1 1
when x = 3, we have {100, 101, 110, 111} - 1 set bit occurs once, 2 set bits occur twice, and 3 set bits occur once, hence 1 2 1
This series is continued. However, summing these up will give me a range, inside of which the answer lies. (Example ans is in [8, 15])
My first solution is the naive approach. Second one is a little mathematical, but not very fruitful.
I was wondering if we could derive a formula any N?
| $F(0) = 0.$
If $2^k \le n \lt 2^{k+1}$, then $F(n) = F(n - 2^k) + F(2^k - 1) + n - 2^k + 1$.
Since $F(2^k -1) = k\,2^{k-1}$, we have $F(n) = F(n-2^k) + k\,2^{k-1} + n - 2^k + 1$.
The recursion works because the numbers between $2^k$ and $n$ all have their highest bit set (those bits give the $n - 2^k + 1$ part of the sum), and the sum of the other bits of those numbers is $F(n - 2^k)$, and the remaining numbers are taken care of by the $F(2^k-1)$ term.
The formula for $F(2^k-1)$ works because each of the $k$ bits of the numbers in ${0, 1,\dots, 2^k - 1}$ is $1$ exactly half of the time.
Edit: Based on Ross Millikan's comment, it is possible to express this as a sum over the bits which are $1$ in $n$, if they are ordered correctly. If ${a_1, a_2,\dots,a_m}$ are the exponents corresponding to the bits which are $1$ in $n$, ordered in increasing order, then $$F(n) = \sum_{i=1}^m a_i\,2^{a_i-1}-i\,2^{a_i}+n+1 = m(n+1) + \sum_{i=1}^m a_i\,2^{a_i-1}-i\,2^{a_i}$$
| {
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Scary looking integral from a movie background I watched recently this movie: https://www.imdb.com/title/tt3149038/mediaviewer/rm261224704
and saw on the poster background, (top-left) the following integral
$$\int_{\large\frac{v}{\sqrt{t}}}^{+\infty}\frac{e^{\large-\frac{u^2}{2}}du}{(1-\tfrac{y^2}{u^2})^{3/2}}$$
I also addded a picture below just in case from the first link it's unclear.
I do not know how to approach such a scary looking integral, elementary methods that I know fails, but I am genuinely curious if it's actually something meaningful or just something trown randomly. Also I learnt normal distribution recently and thought that it might be related, but the denominator doesn't match in any way. How can this integral be approached in order to obtain a closed form and does it have any meaning in mathematics, statistics or physics?
| For the sake of completion, here's a solution to the integral.
We'll analyze the following auxiliary integral:
\begin{align*}
\frac{1}{\sqrt{\pi}}\int_{\sqrt{a^2 +1}}^{\infty} \frac{x e^{- (bx)^2}}{(x^2 -1)^{\frac32}} \mathrm{d}x
& \overset{\color{blue}{t = b^2(x^2 -1)}}{=} \frac{be^{-b^2}}{2\sqrt{\pi}}\int_{(ab)^2}^{\infty} \frac{e^{-t}}{t^{\frac{3}{2}}}\mathrm{d}t\\
& \overset{\text{I.B.P.}}{=} \frac{be^{-b^2}}{\sqrt{\pi}}\left[\frac{e^{-(ab)^2}}{ab} -\int_{(ab)^2}^{\infty} \frac{e^{-t}}{t^{\frac{1}{2}}}\mathrm{d}t\right]\\
& \overset{\color{blue}{s = \sqrt{t}}}{=} e^{-b^2}\left[\frac{e^{-(ab)^2}}{a\sqrt{\pi}} - b\frac{2}{\sqrt{\pi}}\int_{ab}^{\infty} e^{-s^2}\mathrm{d}s\right]\\
& =e^{-b^2}\left[\frac{e^{-(ab)^2}}{a\sqrt{\pi}} - b\,\mathrm{erfc}\left(ab\right)\right]
\end{align*}
Now, using Feynman's trick we can differentiate both sides of the previous equation with respect to $b$ to obtain:
$$
-\frac{2b}{\sqrt{\pi}}\int_{\sqrt{a^2 +1}}^{\infty} \frac{x^3 e^{- (bx)^2}}{(x^2 -1)^{\frac32}} \mathrm{d}x = \left(2b^2 -1 \right)e^{-b^2}\mathrm{erfc}\left(ab\right) -\frac{2be^{-(a^2+1)b^2}}{a\sqrt{\pi}} \tag{1}
$$
Finally, since the integral of interest can be re-written as
$$
\int_{\large\frac{v}{\sqrt{t}}}^{\infty}\frac{e^{-\frac{u^2}{2}}}{\left(1-\frac{y^2}{u^2}\right)^{\frac32}}\mathrm{d}u \overset{\color{blue}{xy =u }}{=} y\int_{\large\color{purple}{\frac{v}{y\sqrt{t}}}}^{\infty} \frac{x^3 e^{-\left(\color{purple}{\frac{y}{\sqrt{2}}} x\right)^2}}{(x^2 -1)^{\frac{3}{2}}}\mathrm{d}x
$$
we can substitute $a = \sqrt{\frac{v^2}{y^2t}-1}$ and $b = \frac{y}{\sqrt{2}}$ into equation $(1)$ to get the final result
$$
\boxed{\int_{\large\frac{v}{\sqrt{t}}}^{\infty}\frac{e^{-\frac{u^2}{2}}}{\left(1-\frac{y^2}{u^2}\right)^{\frac32}}\mathrm{d}u =y^2 e^{-\frac{v^2}{2t}} \sqrt{\frac{t}{v^2 -y^2t}}+\sqrt{\frac{\pi}{2}}\left(1 -y^2\right)e^{-\frac{y^2}{2}}\mathrm{erfc}\left(\sqrt{\frac{v^2}{2t} -\frac{y^2}{2}}\right)}
$$
valid for $v> |y|\sqrt{t}$.
| {
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What is the main difference between pointwise and uniform convergence as defined here? I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.
Let $E$ be a non-empty subset of $\Bbb{R}$. A sequence of functions $\{f_n\}_{n\in \Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if \begin{align}f_n(x)\to f(x),\;\forall\,x\in E.\end{align}
On the other hand $\{f_n\}_{n\in \Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if \begin{align}f_n(x)\to f(x),\;\forall\,x\in E.\end{align}
QUESTION:
Why is $f_n(x)\to f(x),\;\forall\,x\in E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?
| Uniform convergence is actually $\mathcal L^\infty$ convergence, i.e.
$$
f_n \rightrightarrows f [x \in E]\!\! \iff \!\! \sup_{x \in E} \vert f_n - f\vert(x) \to 0[n \to \infty].
$$
This is strictly stronger than pointwise convergence.
Alternatively, uniform convergence implies pointwise convergence, so $f_n \to f$ in both cases.
| {
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How to show that the given set is open? Consider the following subsets of the plane $\mathbb{R}^2$:
$$X=\{(x,y)|y=0\}\cup \{(x,y)|x>0\text{ and}\; y=1/x\}$$
How to show that $A$ and $B$ are open in $X$ under subspace topology.
Efforts:
Let's define $A=\{(x,y)|y=0\}$ and $B=\{(x,y)|x>0\text{ and}\; y=1/x\}$.
To show that $A$ is open I need to find an open set $N$ of $\mathbb{R}^2$ such that $X\cap N=A$. I am not able to proceed further.
I welcome any hints.
Thanks for reading.
| Hint: It might be easier to show that both $A,B$ are closed in $X$, and then since $X = A \cup B$, we immediately have that $A,B$ are both open in $X$.
| {
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Solving the equation $abc=cba$ in the free group. It is known that if two words $a,b$ commute in the free group $F$, then they are powers of the same word, i.e. $a=c^r$ and $b=c^s$, where $c\in F$ and $r,s \in \mathbb Z$.
What happens if there are three words $a,b,c \in F$ such that $a b c=c b a$?
Is there a similar property as above? Or, if not, is there anything that follows from the equation, any information about $a,b$ or $c$? And why?
My goal is to find all solutions for the equation $a b c=c b a$ in the free group of rank two.
| Another way to solve this equation is to replace the product $bc$ with a new variable, $g$ say, and replace $c^{-1}a$ another new variable, $h$ say*. Then your equation becomes
$$
\begin{align*}
abc&=cba\\
c^{-1}abc&=bcc^{-1}a\\
hg&=gh
\end{align*}
$$
Therefore, solutions over the free group $F$ are assignments $(g, h)\rightarrow (w^i, w^j)$ with $i, j\in\mathbb{Z}$ and $w\in F$ such that there is no element $u\in F$ such that $u^k=w$, $k>1$. Substituting in our replacements of $g\leftrightarrow bc$ and $h\leftrightarrow c^{-1}a$, we have all solutions are of the form $$(a, b, c)\rightarrow (vw^i, w^jv^{-1}, v)$$ where $w$ is as above and $v\in F$ is arbitrary.
*This second assignment isn't initially obvious.
| {
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"url": "https://math.stackexchange.com/questions/3063847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
&=H_{1}\left(1-\frac{1}{2}\right)+H_{2}\left(\frac{1}{2}-\frac{1}{3}\right)+H_{3}\left(\frac{1}{3}-\frac{1}{4}\right)\\
&=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{3}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{9}-\frac{1}{12}
\end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
Then,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
| Using summation by parts we have $$\sum_{n\leq N}\frac{H_{n}}{n\left(n+1\right)}=H_{N}\sum_{n\leq N}\frac{1}{n\left(n+1\right)}-\sum_{n\leq N-1}\frac{1}{n+1}\left(\sum_{k\leq n}\frac{1}{k\left(k+1\right)}\right).$$ Clearly $$\sum_{n\leq N}\frac{1}{n\left(n+1\right)}=\left(1-\frac{1}{N+1}\right)$$ then $$\sum_{n\leq N}\frac{H_{n}}{n\left(n+1\right)}=H_{N}\left(1-\frac{1}{N+1}\right)-H_{N}+\sum_{n\leq N-1}\frac{1}{\left(n+1\right)^{2}}.$$ Now, since $H_{N}\sim\log\left(N\right)$ as $N\rightarrow+\infty$, the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
prove $\dim(\operatorname{range}(T)) = \dim(\operatorname{range}(\sqrt{T^*T}))$ I'm a student and I'm studying linear algebra. in Polar Decomposition we have:
for a linear operator $T$, there exist a linear isometry $S$ that:
$$ T =S\sqrt{T^*T}$$
so if $S$ is a linear transformation then it must be $\dim(\operatorname{range}(T)) \leq \dim(\operatorname{range}(\sqrt{T^*T}))$.
But why?
edit: I think it must be equal, I mean:
$$\dim(\operatorname{range}(T)) = \dim(\operatorname{range}(\sqrt{T^*T}))$$
I know that $\dim(\operatorname{range}(T)) = \dim(\operatorname{range}(T^*))$ but because of the square root I cannot prove that $\dim(\operatorname{range}(T)) = \dim(\operatorname{range}(\sqrt{T^*T}))$.
| As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$\dim \text{ran} L =\dim \operatorname{dom} L-\dim\ker L\le\dim \operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
$$
S\big|_{\text{ran} T} :\text{ran} T\to V,
$$ we can see that $\dim \text{ran} (ST)\le \dim \text{ran} T$ as wanted. Moreover, equality holds if $\dim \ker S\big|_{\text{ran}T}=\dim [\ker S\cap\text{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)
To see that $\dim \text{ran}T =\dim \text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $\ker T = \ker (T^*T)$. Now, by dimension theorem, we have
$$
\dim \text{ran} T = \dim V - \dim \ker T = \dim V - \dim \ker (T^*T) = \dim\text{ran}(T^*T).
$$ (Or we can use the fact that $S$ is an isometry.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Showing a statement is true for all positive integers regarding a complex function I am not looking for an answer, for both methods I have one. I'd simply like to check if the statement needs to be proved by induction or simply through rearrangement.
Statement:
For $f(z)=\sum_{n=0}^\infty z^{2^n}$ show that for all positive integers $k$, $f(z)$ satisfies $f(z)=z+z^2+z^4+...+z^{2^{k-1}}+f(z^{2^{k}})$.
My approach:
\begin{align}
z+z^2+z^4+...+z^{2^{k-1}}+f(z^{2^{k}})&=\sum_{n=0}^{k-1} z^{2^n}+\sum_{n=0}^\infty (z^{2^k})^{2^n}\\&=\sum_{n=0}^{k-1} z^{2^n}+\sum_{n=0}^\infty z^{2^{k+n}}\\&=\sum_{n=0}^{k-1} z^{2^n}+\sum_{n=k}^\infty z^{2^n}\\&=\sum_{n=0}^\infty z^{2^n}\\&=f(z)
\end{align}
Would this approach be correct? Or should I prove this statement by induction?
| Personally I would have proved it by induction, since "prove this for all positive integers" or whatever often screams that, but your method is valid as well. (And probably the easier of the two, too.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
A graph-coloring problem where only some of the edges should be bichromatic In a standard graph-coloring problem, it is required that all edges will be bichromatic (i.e., all edges should be connected to two vertices with different colors). What is a term, and some basic references, for a graph-coloring problem in which only a fraction of the edges should satisfy this condition? For example, at least $\frac23$ or at least $\frac12$?
I looked for "fractional coloring" but, apparently, it is a completely different problem.
| A colouring without the condition that edges be bicromatic is called by different names such as relaxed coloring, improper coloring and defective coloring. (warning: the term defective coloring is more commonly used to mean a relaxed coloring with a restriction on maximum degree allowed in colour classes; relaxed coloring with this restriction is explored enough in the literature in both theoretical interest and applications)
What you want is a relaxed coloring with a fraction of edges coloured properly. I remember seeing something similar to this on a paper on .... defective colouring(?), but quite don't remember the source.
I thought this could help someone interested to further explore.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
A function with a non-zero derivative, with an inverse function that has no derivative. While studying calculus, I encountered the following statement:
"Given a function $f(x)$ with $f'(x_0)\neq 0$, such that $f$ has an inverse in some neighborhood of $x_0$, and such that $f$ is continuous on said neighborhood, then $f^{-1}$ has a derivative at $f(x_0)$ given by:
$${f^{-1}}'(x_0)=\frac{1}{f'(x_0)}$$
My questions is - why does $f$ have to be continuous on a whole neighborhood of $x_0$ and not just at $x_0$? Is there some known counter-example for that?
| The continuity condition is not necessary. It's enough that $f$ be injective on some neighborhood. This said, if your function has a sequence of jump discontinuities near $x_0$, you might have that there is no open interval $U$ around $x_0$ for which $f(U)$ is also an interval. This means that $f^{-1}$ might be defined on a strange domain, though we can still technically differentiate it to get the desired result.
Formally, the statement you would need to prove is the following:
Let $A$ and $B$ be subsets of $\mathbb R$ and $f:A\rightarrow B$ and $g:B\rightarrow \mathbb R$. Suppose that $x_0\in A$ is an accumulation point of $A$ and $f(x_0)$ is an accumulation point of $B$. Then,
*
*If two of the derivatives $f'(x_0)$ and $g'(f(x_0))$ and $(f\circ g)'(x_0)$ exist and are non-zero, the third exists as well.
*If all of the derivatives exist, then $(f\circ g)'(x_0)=f'(x_0)\cdot g'(f(x_0)).$
One you have this statement, you can apply it to a pair where we take $g=f^{-1}$. Note that we can make this work even if $f$ isn't defined on an interval around $x_0$ - it's okay as long as we have enough points to define the relevant limit towards $x_0$.
Granted, it is a bit unusual to talk about derivatives on sets that aren't open, but there's no technical limitations preventing it, though the proof of the suggested lemma is a pain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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"answer_id": 3
} |
the inverse of a sum of two symmetric for schur completion? I have a up-triangulate Jacobi matrix J which can be blocked like :
$J = \begin{bmatrix}A & B\\ 0 & C\end{bmatrix} $
both A and C are up-triangulate, we can get Hessian matrix H by:
$H = J'J =\begin{bmatrix}A' & 0\\B'&C'\end{bmatrix} \begin{bmatrix}A & B\\ 0 & C\end{bmatrix} = \begin{bmatrix}A'A & A'B\\B'A & B'B+C'C\end{bmatrix} $
by schur completion, I want to margin $B'B+C'C$, the equation is:
$ A'A - A'B(B'B+C'C)^{-1}B'A $
I know $C^{-1}$ is easy computed because it's a up-triangulate matrix, but unfortunately the add ruin this equation.
I have found that
$ X(X^{-1}+Y^{-1})Y = (X^{-1}+Y^{-1})^{-1} $ only when both X and Y are invertible, which B'B dose not meet
so any one have some brilliant idea to compute this equation easily?
| hi @Omnomnomnom you have given me a better hint to solve this problem , I'd like to list it here
I have checked your equation it's correct except one little mistaken, it should be:
$ (C'C+B'B)^{-1} = (C'C)^{-1} - (C'C)^{-1}B'(I+B(C'C)^{-1}B')^{-1}B(C'C)^{-1} $
I think it's a good solution because B is (small rows * big cols) matrix, we can get
$D = (C'C)^{-1}$
$E = BDB'$
$F = (I+E)^{-1}$
D,E,F are all easy computed, so the result is :
$ (C'C+B'B)^{-1} = D - DB'FBD $
and finally,
$ result = A'A-A'B(B'B+C'C)^{-1}B'A = A'(I-B(B'B+C'C)^{-1}B')A = A'(I-BDB' + BDB'FBDB')A = A'(I-E+EFE)A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Burton Archimedean property proof Can anyone help me understand the following proof of Archimedean property in Burton's Elementary Number Theory book:
Theorem 1.1 Archimedean property. If $a$ and $b$ are any
positive integers, then there exists a positive integer $n$
such that $na \ge b$. Proof. Assume that the statement of
the theorem is not true, so that for some $a$ and $b$, $na < b$ for every positive integer n. Then the set $S = \{b- na | n$ a positive integer$\}$ consists entirely of positive integers. By the Well-Ordering Principle, $S$ will
possess a least element, say, $b- ma$. Notice that $b- (m + 1)a$ also lies in $S$, because $S$ contains all integers of this form. Furthermore, we have $b- (m + 1)a = (b- ma)- a< b- ma$ contrary to the choice of $b - ma$ as the smallest integer in $S$. This contradiction arose out of our original assumption that the Archimedean
property did not hold; hence, this property is proven true.
If he defined $b-ma$ as the least element of the set then how can he define another even lesser element $b-(m+1)a$? Wouldn't that mean that would be a negative number?
| Eventually it means that $b-(m+1)a$ is negative, but the fact that you "know" this is jumping ahead a bit. The assumption is that $na<b$ for every positive integer $n$. That has to include $n=m+1$, so you're assuming $(m+1)a<b$ which is to say $b-(m+1)a >0.$ This leads to a contradiction and then you can conclude that $b-(m+1)a$ is negative. You're letting your intuition get aheas of your logic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Intuitive explanation of solutions to a linear diophantine equation "Given a linear diophantine equation $ax+by=c$ with a particular solution $(x_0,y_0)$ the general solution is given by $$\biggl(x_0-\frac{b}{gcd(a,b)}t,y_0+\frac{a}{gcd(a,b)}t\biggr)$$ for all $t\in \mathbb{Z}$"
I understand the proof of this theorem pretty well but would appreciate an intuitive explanation of why this general solution gives all the solutions to the equation...
| Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.
Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.
But, sort of as a dufus, I could insist on exchanging more bills.
So if I give you $x_0 + \frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - \frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$\frac{ab}{gcd(a,b)}t - \frac{ba}{gcd(a,b)} t = 0$$
So we would still exchange exactly $c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solving $\lim\limits_{x\rightarrow\infty} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}$ To investigate the convergence of a series I have to solve the folliwing limit:
\begin{equation}
\lim\limits_{x\rightarrow\infty} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}
\end{equation}
It should be $\frac{1}{2}$ but i can't quite seem to get to that solution. I've tried to factor the square root out of the quotient which was:
\begin{equation}
\lim\limits_{x\rightarrow\infty} \sqrt{\frac{((x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1})^2}{x^2(x + 1)}}
\end{equation}
Then i worked out the square in the numerator which was:
\begin{equation}
(x^2-1)^2 (x + 2)-2x^2(x^2-1)\sqrt{x+1}\sqrt{x+2}+x^4(x+1)
\end{equation}
I could then factor the terms and take and divide the numerator with the (x+1) from the denominator. Then i could expand the terms in the numerator wich became:
\begin{equation}
\lim\limits_{x\rightarrow\infty}\sqrt{\frac{(x^4+x^3-3x^2-x+2)-\sqrt{16x^8+32x^7-12x^6-20x^5+8x^4}+x^4}{x^2}}
\end{equation}
Now i can take the $16x^8$ out of the root and then i then looked at the terms with the highest exponent so i had:
\begin{equation}
\lim\limits_{x\rightarrow\infty}\sqrt{\frac{x^4-4x^4+x^4}{x^2}} =\lim\limits_{x\rightarrow\infty}\sqrt{\frac{-2x^2}{x^2}}
\end{equation}
Which could only be solved with complex numbers, so i should be wrong somewhere in my calculations, since i know that i should get $\frac{1}{2}$. I also checked my sollution with WolframAlpha which also gave $\frac{1}{2}$ so I know that the sollution i have is correct.
Would anyone know where i was wrong or how i could better solve it?
| Another way:
\begin{eqnarray*} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}
& = & \frac{(x-1)(x+1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}} \\
& = & \frac{(x-1)\sqrt{(x+1)(x + 2)}-x^2}{x} \\
& \stackrel{x^2 = x(x-1)+x}{=} & \underbrace{\frac{x-1}{x}}_{\stackrel{x \to +\infty}{\longrightarrow}1}\underbrace{\left(\sqrt{(x+1)(x + 2)} - x\right)}_{= \frac{3x+2}{\sqrt{(x+1)(x + 2)} + x}\stackrel{x \to +\infty}{\longrightarrow}\frac{3}{2}} - 1 \\
& \stackrel{x \to +\infty}{\longrightarrow} & 1 \cdot \frac{3}{2} -1 = \frac{1}{2}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065147",
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"question_score": "4",
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Evaluate the limit $\lim_{n\to\infty}\log_a\left(\frac{4^nn!}{n^n}\right)$
Evaluate the limit:
$$
\lim_{n\to\infty}\log_a\left(\frac{4^nn!}{n^n}\right)\\
a>0\\
a \ne 1
$$
I've started with defining another sequence. Let:
$$
y_n = a^{x_n} = \frac{4^nn!}{n^n}
$$
Consider the fraction:
$$
\frac{y_{n+1}}{y_n} = \frac{4^{n+1}(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{4^nn!}\\
= \frac{4n^n}{(n+1)^n}
$$
Consider the limit:
$$
\begin{align}
\lim_{n\to\infty}\frac{y_{n+1}}{y_n} &= \lim_{n\to\infty}\frac{4n^n}{(n+1)^n} \\
&= \lim_{n\to\infty}4\left(\frac{n}{n+1} \right)^n \\
&= {4\over e} > 1
\end{align}
$$
So by this $y_n$ is divergent. Which means:
$$
\lim_{n\to\infty}y_n = \infty
$$
Now I'm having difficulties translating it in a backward direction. We have that:
$$
\lim_{n\to\infty}y_n = \lim_{n\to\infty}a^{x_n} = \infty
$$
Or:
$$
\log_a \lim_{n\to\infty}a^{x_n} = \log_a(\infty)
$$
The answer suggests that:
$$
\lim_{n\to\infty}x_n =
\begin{cases}
+\infty,\ a > 1\\
-\infty,\ 0 < a < 1
\end{cases}
$$
And I don't see where this appears when going backward from $a^{x_n}$ to $x_n$. Could you please explain that to me?
| Hint: Note that $$\log_a{x}=\frac{\ln(x)}{\ln(a)}$$
| {
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There do not exist $m\times n$ and $n\times m$ rectangular matrices $A$ and $B$ . t. $AB=I_m$ (The $m\times m$ Identity matrix)
Let $m,n\in\Bbb N$ be positive with $m>n$. Prove that $\nexists A\in M_{mn}\ \&\ B\in M_{nm},\ AB=I_m$ (The $m\times m$
Identity matrix)
I have noticed that this is not true if the matrices are multiplied the other way. That is, say $B=\begin{bmatrix}
1 &0 &0 \\
0& 1& 0
\end{bmatrix}$ and $A=\begin{bmatrix}
1 &0 \\ 0&1 \\ 0&0 \end{bmatrix}$
Then the product $BA$ yields $I_2$, But not the other way. Here this does not contradict with the statement in the question because of the order of the matrix mentioned in the question (that is the one with a higher number of rows should be multiplied first).
So, a proof for the given question is highly appreciated
| Let $B$ be a matrix with more columns $B_1,\ldots, B_n$ than rows. Then there exists a $l$ s.t. $B_l = \sum_{i \not = l} c_iB_i$, where the $c_i$s are scalars.
However, if there is a matrix $A$ such that $AB = I$ (so $A$ has $m$ rows) then for each $k$, the following must hold: $A_k \cdot B_k = 1$ and $A_k \cdot B_i = 0$ for each $i \not = k$. Thus, letting $A_l$ be the $l$-th row of $A$, the following must hold: $A_l\cdot B_l = 1$ and $A_l \cdot B_i = 0$ for all $i \not = l$. However, $A_l \cdot B_i = 0$ for all $i \not = l$ would imply $A_l \cdot B_l = \sum_{i \not = l} c_iA_l \cdot B_i = 0$, contradicting the equation $A_l \cdot B_l = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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quotient of normalizer and centralizer is cyclic group It's known that if $G/Z(G)$ is a cyclic group, then $G$ is Abelian. Since $G/Z(G)$ is just the special case $H=G$ in the $N/C$ theorem $C_G(H)\triangleleft N_G(H)$. I wonder if the below statament is true:
If $H$ is a subgroup of $G$, if $N_G(H)/C_G(H)$ is a cyclic group, then $H$ is Abelian.
I tried using the proof in the original statement, but it didn't work out. If we set $N_G(H)/C_G(H)=\langle aC_G(H)\rangle$, how can i represent an element in $H$?
| Hint: $H \cap C_G(H)=Z(H)$ and $H \unlhd N_G(H)$
| {
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A finite summation of double binomial coefficients I find the following identity and have checked on Mathematica, while I have no idea how to prove it:
$$\sum_{j=0}^n(-1)^{n-j}\binom{p+j}{j}\binom{n+\beta}{n-j}=\binom{p-\beta}{n}, \quad \beta>-1, \quad p>\beta-1.$$
It seems that proof by induction does not work and because of the gamma function, the transform technique does not work either.
| We have
$$
\eqalign{
& \sum\limits_{j = 0}^n {\left( { - 1} \right)^{\,n - j} \left( \matrix{
p + j \cr
j \cr} \right)\left( \matrix{
n + \beta \cr
n - j \cr} \right)} = \cr
& = \sum\limits_{j = 0}^n {\left( { - 1} \right)^{\,n} \left( \matrix{
- p - 1 \cr
j \cr} \right)\left( \matrix{
n + \beta \cr
n - j \cr} \right)} = \cr
& = \left( { - 1} \right)^{\,n} \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{
- p - 1 \cr
j \cr} \right)\left( \matrix{
n + \beta \cr
n - j \cr} \right)} = \cr
& = \left( { - 1} \right)^{\,n} \left( \matrix{
- p - 1 + n + \beta \cr
n \cr} \right) = \cr
& = \left( \matrix{
p - \beta \cr
n \cr} \right)\quad \left| \matrix{
\;n \in Z \hfill \cr
\;\forall p,\beta \hfill \cr} \right. \cr}
$$
where the steps are:
- upper negation (always valid for integer $j$);
- we can omit summming limits, because they are implicit in the two binomials;
- convolution;
- upper negation (always valid for integer $n$).
| {
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Do the axioms of ordered field imply that $a\cdot 0=0$ and $0<1$? We introduce some definitions:
A structure $\mathfrak{A}=\langle A,<,+,\cdot,0,1 \rangle$ where $<$ is a linear ordering, $+$ and $\cdot$ are binary operations, and $0,1$ are constants such that all properties 1-12 are satisfied is called an ordered field.
For all $a,b,c\in A$:
*
*$a+b=b+a$.
*$(a+b)+c=a+(b+c)$.
*$a+0=a$.
*There exists $a'\in A$ such that $a+a'=0$. We denote $a'=-a$, the opposite of $a$.
*$a<b\implies a+c<b+c$.
*$a\cdot (b+c)=a\cdot b + a\cdot c$.
*$a\cdot b = b\cdot a$.
*$(a\cdot b)\cdot c = a\cdot (b\cdot c)$
*$a\cdot 1=a$
*For $a\neq 0$, there exists $a'\in A$ such that $a\cdot a'=1$. We denote $a'=a^{-1}$, the reciprocal of $a$.
*$a<b$ and $0<c$ $\implies a\cdot c < b\cdot c$.
*$0\neq 1$
In the familiar structure $\mathfrak{R}=\langle \Bbb R,<,+,\cdot,0,1 \rangle$ where $\Bbb R$ is the set of real numbers, it is trivial that $a\cdot 0=0$ and $0<1$.
I would like to prove that $a\cdot 0=0$ and $0<1$ hold for any ordered field. I have tried to no avail. Naturally, a question arises in my mind.
Do the axioms of ordered field 1-12 imply that $a\cdot 0=0$ and $0<1$?
Please shed me some light. Thank you for your help!
| $a \cdot 0=0$ is a consequence of the ring axioms: $a \cdot 0= a \cdot (0+0)= a \cdot 0+ a\cdot 0$.
Assume $0 > 1$. Then $-1 = 0 + (-1) > 1 + (-1) =0$. Therefore, for any $a < b$, $(-1)a < (-1)b$, hence $0 = 1 \cdot a + (-1)a = a + (-1)a < b + (-1)a < b + (-1)b= 1 \cdot b + (-1)b = 0$, a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\gcd(a,b)=1, x^a = y^b\Rightarrow x = n^b$, $ y = n^a$ for an integer $n$. If $ a$, $ b$, $ x$, $ y$ are integers greater than $1$ such that $ a$ and $ b$ have no common factor except $1$ and $ x^a = y^b$ show that $ x = n^b$, $ y = n^a$ for some integer $ n$ greater than $1$.
I started this way: $x^a=y^b$ $=>$ $x=y^{b/a}=(y^{1/a})^b$. Thus $x$ is an integer, and so is $y^{1/a}$, so $y=n^a$ for some integer $n$ greater than $1$. Then from the given relation, we get $x=n^b$.
I am not very convinced with my solution, as I feel I did some mistake. Can anyone point out where the mistake is? Any other solution is also welcome :)
| Here is an alternate method.
Note that set of primes dividing $x$ and $y$ are same. Take any prime $p$ diving $x$(and
hence $y$), and let $\alpha$ is the maximum power of $p$ in $x$ and $\beta$ is the maximum power of $p$ in $y$. Then $x^a=y^b \implies p^{\alpha a}=p^{\beta b}$, which implies $a|\beta b $ and $b| \alpha a$. But remember that $gcd(a,b)=1$. So, $a|\beta$ and $b|\alpha$.
Suppose, $\beta= a\cdot \beta_p$ and $\alpha=b\cdot \alpha_p$. Then we have, $\alpha a=\beta b$ or, $b\alpha_pa =a\beta_p b$ or, $\alpha_p=\beta_p$. So, for each prime $p$ diving $x$, we have such $\alpha_p$. Check that $n=\prod_{p|n}p^{\alpha_p}$ satisfies the required property.
| {
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Counting measure in integral. Let $\mu$ counting measure on $X$ countable set.
This is correct?
$\int_{X}|f|^pd\mu=\int_{\bigcup_{k\in\mathbb{N}} x_k} |f|^pd\mu=\sum_{k\in\mathbb{N}}\int_{x_k}|f|^pd\mu=\sum_{k\in\mathbb{N}}\int |f|^p\mathcal{X}_{x_k}d\mu=\sum_{k\in\mathbb{N}} |f(x_k)|^p\int \mathcal{X}_{x_k}=\sum_{k\in\mathbb{N}} |f(x_)|^p\mu(x_k)=\sum_{k\in\mathbb{N}}|f(x_k)|^p$
Therefore ${L(X)}^p={l(X)}^p$
| The equalities are in principle correct provided that $X$ is not finite.
Not completely correct is the notation that is practicized.
Every $x_k$ in the equalities should be replaced by $\{x_k\}$.
Further IMV it is a bit overdone.
If $\mu$ denotes the counting measure and $X$ is countable then you can write immediately:$$\int_X|f|^pd\mu=\sum_{x\in X}|f(x)|^p$$
There is no need to go for $\mathbb N$ as index set. It is even wrong if $X$ is a finite set.
Where it concerns $L(X)^p=l(X)^p$ see the comment of Kavi to your question.
| {
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Find $a,b$ such that $\mathbb R^3 = V \oplus W$ I have some doubts with this task:
Let $a,b \in \mathbb R$. In linear space $\mathbb R^3$ we have:
$$V = span([1,1+a,-2]^T,[2,6,-2-a]^T)= span(v_1,v_2) $$ and $$W = span([0,3,-1-b]^T,[2,2+b,-2]^T) = span(w_1,w_2)$$
Find $a,b$ such that $\mathbb R^3 = V \oplus W$
If $\mathbb R^3 = V \oplus W$ then $V \cap W = \left\{\vec{0}\right\}$
I think that I should consider these cases:
$1.$ $(v_1,v_2,w_1)$ are linear independent and $w_2 = \alpha w_1 $
$2.$ $(v_1,v_2,w_2)$ are linear independent and $w_1 = \alpha w_2 $ (but it is equivalent to $1$ so I don't have to check that)
$3.$ $(w_1,w_2,v_1)$ are linear independent and $v_2 = \alpha v_1 $
$4.$ $(w_1,w_2,v_2)$ are linear independent and $v_1 = \alpha v_2 $ (but it is equivalent to $3$ so I don't have to check that)
but there are a lot of calculus and I am not sure if (i) this is correct way (ii) how to do this quickly
Can somebody help me with this problem? Thanks for your time.
| Observe that both $V$ and $W$ can be at most two dimensional each. If both are two dimensional then their intersection cannot be just $\{0\}$. If one of them is $0$ dimensional OR both are $1$ dimensional then $\Bbb{R}^3$ cannot be their direct sum. So the only two possibilities are:
*
*$V$ is $1-$dim and W is $2-$dim.
*$V$ is $2-$dim and W is $1-$dim.
In the first case $v_1$ and $v_2$ must be dependent vectors and $v_1,w_1,w_2$ must be independent vectors. That means
$$k\begin{bmatrix}1\\1+a\\-2\end{bmatrix}=\begin{bmatrix}2\\6\\-2-a\end{bmatrix} \implies k=2 \text{ and } a=2$$
For the independence of $\{v_1,w_1,w_2\}$, the following matrix should have full rank.
$$\begin{bmatrix}1&3&-2\\0&3&-1-b\\2&2+b&-2\end{bmatrix} \longrightarrow \begin{bmatrix}1&3&-2\\0&3&-1-b\\0&b-4&2\end{bmatrix} \longrightarrow \begin{bmatrix}1&3&-2\\0&3&-1-b\\0&0&\frac{(b-2)(b-1)}{3}\end{bmatrix}.$$
Thus for full rank, we want $b \neq 1,2$.
Thus with $a=2$ and $b \neq 1,2$, $\Bbb{R}^3=V \oplus W$.
Similarly you can do the other case.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove surjectivity of determinant It may seem obvious that the determinant of a square matrix as a map is surjective (since there is always a choice of matrix entries that yields a real number). I can't prove this statement. Any clue please?
| Hint:
Given $r\in \Bbb R $, take the diagonal matrix
$\begin{bmatrix}r&0&\cdots & 0\\
0&1&\cdots & 0\\
\vdots&\vdots&\ddots&\vdots\\0 & 0 & \cdots &1\end{bmatrix} $.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence of $\int_1^2 \frac{1}{\sqrt \ln x} \mathrm { d}x$ by Asymptotic Comparison Test. According to the material I have,
If $f:[a,b)\to \mathbb{R} ,b\in \mathbb{R},f\geq0 $
Then:
i) If $f$ is an infinite function of real order $\alpha\lt1$ with respect to $\frac {1}{x-b}$ $\implies$ $\int_a^bf(x)\mathrm { d}x\lt\infty$
ii) If $f$ is an infinite function of real order $\alpha\geq1$ with respect to $\frac {1}{x-b}$ $\implies$ $\int_a^bf(x)\mathrm { d}x=\infty$
Based on this information I am trying to evaluate the convergence of $\int_1^2 \frac{1}{\sqrt \ln x} \mathrm { d}x$
In order to determine the order of my function I consider $$\lim_{x\to1+}{\frac{1}{\sqrt\ln x \over\frac{1}{(x-2)^\alpha}}}=\lim_{x\to1+}\frac {(x-2)^\alpha}{\sqrt\ln x}$$
Despite trying various methods I haven't been able to obtain a real number as a limit value for some $\alpha\in\mathbb{R}$
| An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $f\colon (a,b]\to\mathbb{R}$ and look at the $a=1$ end.
Rewrite, for $x>1$,
$$
\sqrt{\ln x} = \sqrt{\ln(1+(x-1))}\,.
$$
This seems silly, but now, recall that
$$
\lim_{u\to 0} \frac{\ln(1+u)}{u} = 1
$$
so that
$$
\lim_{x\to 1} \frac{\sqrt{\ln x}}{(x-1)^{1/2}} = \lim_{x\to 1} \sqrt{\frac{\ln x}{x-1}}= \lim_{x\to 1} \sqrt{\frac{\ln(1+(x-1))}{x-1}} = \sqrt{\lim_{x\to 1} \frac{\ln(1+(x-1))}{x-1}} = \sqrt{1}=1
$$
Can you conclude?
| {
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Does an element of a set, that can't be in a list, make that set uncountable? We can map the infinite subsets of $\mathbb N$ to the finite subsets of $\mathbb N$
The finite subset will be a prefix, of the infinite subset (that it is paired with), that has not yet been used.
$1 \mapsto \{ \color{red}{1} ,2,3,4,5,6,7,8 \dots \} \mapsto \{ \color{red}{1} \}$
$2 \mapsto \{ \color{red}{2} ,4,6,8,10,12,14 \dots \} \mapsto \{ \color{red}{2} \}$
$3 \mapsto \{ \color{red}{1,3} ,5,7,9,11,13,15 \dots \} \mapsto \{ \color{red}{1,3} \}$
$4 \mapsto \{ \color{red}{1,2} ,3,7,9,19,27,31 \dots \} \mapsto \{ \color{red}{1,2} \}$
$5 \mapsto \{ \color{red}{1,2,3} ,4,21,22,25,32 \dots \} \mapsto \{ \color{red}{1,2,3} \}$
$6 \mapsto \{ \color{red}{2,3} ,4,6,7,8,21,55,58 \dots \} \mapsto \{ \color{red}{2,3} \}$
$7 \mapsto \{ \color{red}{2,3,4} ,6,7,8,9,21,55,58 \dots \} \mapsto \{ \color{red}{2,3,4} \}$
$8 \mapsto \{ \color{red}{2,3,4,6} ,7,9,21,55,58 \dots \} \mapsto \{ \color{red}{2,3,4,6} \}$
$9 \mapsto \{ \color{red}{2,3,4,6,7} ,6,7,8,21,55,58 \dots \} \mapsto \{ \color{red}{2,3,4,6,7} \}$
$\dots$
We can then find an infinite set that is not in this list (by diagonalization).
$N \mapsto \{ \color{red}{4,5,8,9,\dots} \} \mapsto \{ \color{red}{4} \} \lor \{ \color{red}{4,5} \} \lor \{ \color{red}{4,5,8} \} \lor\{ \color{red}{4,5,8,9} \} \dots$
We can then pair our infinite set, that can't be in our list, with $\{ \color{red}{4} \}$ if it is not been used or $\{ \color{red}{4,5} \}$ or $\{ \color{red}{4,5,8} \}$ etc... until we find some finite set that has not been used.
Since we need sets with a finite number of elements, and we have sets with an infinite number of elements to choose from, we will always find a finite prefix from each infinite set.
So, if we can find a finite set for every infinite set IN or NOT IN our list, and our set of finite sets is countable then can we still say that our set of infinite sets is uncountable based on there being an element that can't be in a list?
| Even though that specific infinite subset isn't in our list, we can still have all its finite prefixes in our list of finite subsets.
For instance, say the 10th infinite subset in your list is $\{4, \ldots\}$. Then the 20th is $\{4, 5, \ldots\}$. Then the 30th is $\{4, 5, 8, \ldots\}$. And so on. None of them have to be your infinite set, all of them can be different from your infinite set early enough to allow your set to be the result of the diagonalization. But all of your set's prefixes are in our list of finite subsets.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} * \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$ to $\frac{\sqrt{mnc}}{a^9cmn}$ I need to simplify $$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$$
The solution provided is: $\dfrac{\sqrt{mnc}}{a^9cmn}$.
I'm finding this challenging. I was able to make some changes but I don't know if they are on the right step or not:
First, I am able to simplify the left fractions numerator and the right fractions denominator:
$\sqrt{mn^3}=\sqrt{mn^2n^1}=n\sqrt{mn}$
$\sqrt{m^2c^4}=m\sqrt{c^2c^2} = mcc$
So my new expression looks like:
$$\frac{n\sqrt{mn}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{mcc}$$
From this point I'm really at a loss to my next steps. If I multiply them both together I get:
$$\frac{(n\sqrt{mn})(a^{-7}n^{-2})}{(a^2\sqrt{c^{-3}})(mcc)}.$$
Next, I was thinking I could multiply out the radical in the denominator but I feel like I need to simplify what I have before going forwards.
Am I on the right track? How can I simplify my fraction above in baby steps? I'm particularly confused by the negative exponents.
How can I arrive at the solution $\dfrac{\sqrt{mnc}}{a^9cmn}$?
| Use
$$a^{-b}=\frac{1}{a^{b}} .$$
So
\begin{align}
\frac{(n\sqrt{mn})(a^{-7}n^{-2})}{(a^2\sqrt{c^{-3}})(mc^2)}&=\frac{n\sqrt{mn}}{(a^2{c^{-3/2}})(mc^2)(a^{7}n^{2})}\\
&=\frac{(\sqrt{mn})c^{3/2}}{(mc^2)(a^{9}n)}\\
&=\frac{\sqrt{mnc}}{a^{9}cmn}\\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\lim_{x\to 0} \frac {(\sin(2x)-2\sin(x))^4}{(3+\cos(2x)-4\cos(x))^3}$? $$\lim_{x\to 0} \frac {(\sin(2x)-2\sin(x))^4}{(3+\cos(2x)-4\cos(x))^3}$$
without L'Hôpital.
I've tried using equivalences with ${(\sin(2x)-2\sin(x))^4}$ and arrived at $-x^{12}$ but I don't know how to handle ${(3+\cos(2x)-4\cos(x))^3}$. Using $\cos(2x)=\cos^2(x)-\sin^2(x)$ hasn't helped, so any hint?
| Hint: Note that
$$ 3+\cos(2x)-4\cos(x) = 3 + 2\cos^2(x) - 1 - 4\cos(x) = 2(\cos(x)-1)^2, $$
and that
$$ \sin(2x) - 2\sin(x) = 2\sin(x)\cos(x)-2\sin(x) = 2\sin(x)(\cos(x)-1). $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Similarity of two self-adjoint operators I am wondering whether the following is correct:
Let $A$ and $B$ be two bounded self-adjoint, positive and invertible linear operators such that $\sigma(A)=\sigma(B)$ and $AB=BA$. Can we say that $A$ is necessarily similar to $B$?
I just couldn't find an answer.
Thanks.
Math.
| Even if $\sigma(A)=\sigma(B)$, similarity sees multiplicity. For instance let
$$
A=\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&2\end{bmatrix},\ \ \ B=\begin{bmatrix} 1&0&0\\ 0&2&0\\0&0&2\end{bmatrix}.
$$
Then $AB=BA$, $\sigma(A)=\sigma(B)=\{1,2\}$, but they are not similar (for instance, because they don't have the same trace).
| {
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"timestamp": "2023-03-29T00:00:00",
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Diffeomorphism between the unit ball to itself Let $T:B\to B$ be a diffeomorphism , where $B\subset\mathbb{R}^n$ is the unit ball.
I want to show there exists $x\in B$ such that $|J_T(x)|=1$ ($|J_T|$ is the absolute value of the Jacobian of $T$)
Intuitively speaking, since $T$ maps the ball to itself, I`d expect that $|J_T(x)|≡1$, Since the map does not inflate the ball, so it might only perform rotations or reflections. However my intuition here must be false.
My line of reasoning was as following:
$Volume(B):=\int_B1dt=\int_{T^-1(B)}|J_T(x)|dx=\int_B|J_T(x)|dx$
Where the first equality is how we defined volume, the second is by changing variables (Possible since $T$ is diffeomorphism) and the last equality because $T$ maps $B$ to itself, so $T^{-1}$ must also map $B$ to itself.
From here one can conclude that either $|J_T(x)|≡1$ or that if $|J_T(x)|\neq1$ (Say $|J_T(x)|<1$ for example) there must exist a point where $|J_T(x)|>1$ or we`d get $Volume(B)>Volume(B)$ as a contradiciton.Now simply using continuouity of $|J_T|$ along with the I.V.T gets the desired result.
From this proof, it seems like there might be such a map that isn`t a combination of rotations or reflections, and that $T$ might also "inflate" areas of the ball. This is very counter-intuitive to me, and I could not imagine such mapping. I'd be glad if someone could provide with an example, and perhaps some better intuition on the behaviour of diffeomorphisms.
If such an example does not exist, I'd love to see a proof.
| One possible diffeomorphism from the unit ball in $\mathbb{R}^2$ to itself, which isn't rigid in the way you expect, could be described in polar coordinates as
$$(r, \theta) \mapsto \left(\frac{r + r^3}{2}, \theta \right),$$
or in rectangular coordinates as
$$(x, y) \mapsto \frac{1 + x^2 + y^2}{2} (x, y).$$
(The first representation was what I came up with to make it easy to show the function is bijective, whereas the second representation makes it easier to show the differentiability of the function and its inverse at the origin.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Converting standard form to vertex form, parental homework help I'm trying to help my son with his homework but am having trouble feeling confident that I know what the assignment is asking for.
I've been learning (maybe relearning) about standard vs vertex form quadratic equations.
The first part of the homework states:
Rewrite the standard form $f(x) = ax^2 + bx + c$. Once you've done that set the vertex form equal to zero and solve for x. Show your work:
All of the examples I can find on how to convert a standard form to vertex are using actual numbers for $a$ and $b$. Because the steps involve converting to a perfect trinomial, I'm not sure how to "rewrite" the equation without using actual values as an example.
It then asks:
You've now discovered the ___ ____! Use it to solve this: $x^2 + 4x - 11 = 0$
I'm not sure what the underlines are asking for - "perfect trinomial"? I can figure out how to the solve the equation but I want to be sure I know how using the expected method.
| This is not entirely straightforward, but we'll walk through it.
You start with an equation in standard form: $y = ax^2 + bx + c$. To convert this to 'vertex form' we must complete the square.
$$y = ax^2 + bx + c$$
$$y - c = ax^2 + bx$$
$$y - c = a(x^2 + \frac{b}{a}x)$$
$$y - c + a(\frac{b^2}{4a^2}) = a(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}) = a (x + \frac{b}{2a})^2$$
Hence we get,
$$y = a(x + \frac{b}{2a})^2 + (c - \frac{b^2}{4a})$$
If unfamiliar with the process of completing the square as above see here. (You can also always multiply this out and check that it is, indeed, the same as $y = ax^2 + bx + c$).
Now, the problem is telling you to set this equal to 0 and solve for $x$. It is much easier to solve for $x$ from vertex form (hence the conversion) as we will see:
$$0 = a(x + \frac{b}{2a})^2 + (c - \frac{b^2}{4a})$$
$$(x + \frac{b}{2a})^2 = \frac{-c + \frac{b^2}{4a}}{a} = \frac{-4ac + b^2}{4a^2}$$
Take the square root of both sides and we get,
$$x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
which becomes the familiar quadratic equation:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Can Someone Explain the Set $\{ ∈ ℝ^ ∣ d(x,0) = 1\}$? I'm trying to understand this set. I understand most of the individual parts such as $∈$ means "an element of", $ℝ$ is all real numbers, and $d(x,0) = 1$ is the distance from $x$ to $0$, as in absolute value.
I don't understand what $x$ means in this situation though. Is it considered a variable? Is it multiple values?
I also don't understand what the actual elements of the set are. Is the value of $x$ and element in the set? Also, are there any other numbers with an absolute value of $1$ other than $1$ and $-1$? If not, are those the only possible values for $x$?
In this situation, $N$ is a number of dimensions, but I don't understand how that relates to real numbers, nor what real numbers to any power would be.
Thanks.
| The notation $d(x,0)=1$ seems to indicate that this is a problem in a general metric space. This is not necessarily the absolute value, as there can be more abstract metrics. This set is the unit ball in $\mathbb{R}^N$ with respect to this metric.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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